m 


c* 


-SO 


REESE    LIBRARY 

Of    THK 

UNIVERSITY   OF   CALIFORNIA. 

Received. s^^L^Zl^  rSS(/ 

Accessions  No.2*£~&fi*^     Shelf  No 

* £ 


KEY 


TO     ROBINSON'S 


UNIVERSITY   ALGEBRA; 


CONTAINING,    ALSO, 


A  SHORT  TREATISE  ON  THE  INDETERMINATE 
AND  DIOPHANTINE  ANALYSIS. 


AND 


SOME   MISCELLANEOUS    EXAMPLES, 


JBesffltie'O  for  5Teacf)ers  attfi  SatuUents. 


NEW    YORK: 

IVISON  &   PHINNEY,    321   BROADWAY 

CINCINNATI:      JACOB      ERNSf 
CHICAGO :   8.  C.  GRIGGS  &  CO.,  39  &  41  LAKE  ST. 

ST.  LOUIS :   KEITH  k  WOODS.   BUFFALO  :    PHINNEY  &  CO. 

185  9. 


■ft 


Entered  according  to  act  of  Congress,  in  the   year  1847,  by 

HORATIO   N.   ROBINSON, 

in  the  Clerk's   Office  of  District  Court   of  the  District  of  Ohu* 


\^V       OF  THE  '       \\ 

'UNIVERSITY' 

flfe 


A  Key,  to  a  mathematical  work,  is  very  proper  in 
its  place  ;  but  to  be  constantly  at  hand,  and  consulted 
too  often,  might  prove  injurious :  we  must  not,  how- 
ever, confound  the  improper  use  of  a  thing  with  the 
thing  itself.  Those  who  condemn  keys,  in  general 
terms,  should  condemn  teachers  also ;  for  a  key  is 
neither  more  nor  less  than  a  teacher,  in  another  shape. 

The  self-taught  are  generally  sound  and  vigorous ; 
but  if  they  disregard  the  works  and  teachings  of  others, 
they  will  be  found  to  be  wanting  in  that  certain  sym- 
metry and  polish  of  mind,  so  characteristic  of  educated 
men. 

So  it  is  with  an  algebraist ;  h'e  may  go  through  his 
text-books,  solve  every  problem,  independent  of  all 
external  aid,  and  if  he  does  not  compare  his  work 
with  the  works  of  others,  he  cannot  know  whether  he 
is  skillful  or  otherwise ;  for  it  is  only  by  comparison 
that  we  measure  excellence.  No  solution  of  a  prob- 
lem, or  of  an  equation,  should  be  called  good,  if  better 
can  be  found ;  hence  it  is  important  that  more  than 
one  standard  of  attainment  should  be  before  the  pupil ; 
and  those  who  really  become  eminent,  in  any  science, 
are  those  whose  talents  and  dispositions  enable  them 
to  gather  knowledge  from  every  possible  source. 


PREFACE  TO  REVISED  EDITION. 


In  1857  the  author  and  publisher  thought  it  advisable 
to  enlarge  the  University  Algebra  by  some  twenty-four 
pages,  and  some  more  than  twenty  additional  examples. 

Solutions  of  these  are  inserted  subsequent  to  page  76. 
Also,  solutions  of  several  other  problems  are  inserted, 
which  were  omitted  in  the  former  editions  of  this  work. 


'UNIVERSITY 
KEY 


TO 

ROBINSON'S   ALGEBRA 


SECTION     II. 

CHAPTER  I. 

EQUATIONS. 

None  of  the  questions  in  this  chapter  require  the  aid  of 
a  key,  until  we  come  to  the  15th,  page  65. 

(15.)  (4x—4a      \4     I6x—I6a    4a 


/4x—4a      \4_J 
\      3  /3~~ 


=    his 


9  3 

stock  at  the  commencement  of  the  third  year,  before  his  ex- 
penses are  taken  out. 

tj  /16a? — 16a    4a      \4 

Hence,  ( a  }_=2a; 

V        9  3/3 

Reduced  gives  x=  14800,  Am. 

(16.)  Put  a=99,  #=time  past.     Then  a — x==  time  to 
come,  and  per  question, 

2a?    4a — 4x 


3  5 

(17.)  Let  x=  the  whole  composition. 
Then  per  question, 


z=54. 


KEY  TO  ALGEBRA. 

2a?  , 

r-10=nitre. 

3 

x 

4£  =sulphur. 

6 

2x     10 

— -f— — 2= charcoal. 
21      7 


2x     x     2x  10 

By  addition,    — +-- f h3H =x 

J  3      6     21  7 

Multiply  by  6,  and  drop  5a;  from  both  sides,  and  we  have 

4x  60 

7  7 

or,  4#+21-7+60=7#    .  •  .  .#=69. 


(18.)  Puta=183  ;  #=what  the  1st  received  ;  then 

a — x=2d  received. 

4a?     3a — 3a? 

Per  question,  — = #=63. 

7  10 

(19.)  Put  a=68,  #=  the  greater  part,  and  a — x=  the 

less.  84— #=3(40— a+  x) #=42. 

(20.)  The  distance  from  A  to  B  put  =2#. 

The  distance  from  C  to  D  "    =3x. 

Then,  3  times  the  distance  from  B  to  C  must  be 

x    Sx  ,  xx 

— | or  the  distance  is,   — | — 

2      2  6     2 

x    x 
Hence  the  whole  distance  is,      5x-\ 1 — =34. 

6     2 
(21.)  Let  x= the  flock. 

2x 
The  first  party  left  him 6. 

x 

The  second  left     3—10=2. 

3 


EQUATIONS.  7 

(23  )  Observe  that  for  every  vessel  he  broke  he  lost  12 
cents :  3  cents  fee  and  9  cents  forfeiture. 

300—12^=240  ...  - #=5. 

(24.)  Had  he  not  been  idle  he  would  have  been  entitled 
to  ab  cents.  But  he  was  idle  x  days  at  a  loss  of  (b-\-c) 
cents.     Hence,  ab — (b-\-c)x=d. 

ab — d 


b+c 


(25.)  Put  5x= 
Then  a — 5x = 

=  less  part 
the  greater  part. 

Per  question, 

3 

a — 7x = 2  Ox (a — 5x) 

or, 
or, 
Therefore, 

7a—49x=  1 40#— 3a+ 1  Sx 

204.r=10a=10-204 

#=10 

5^=50=  the  less  part. 

(26.)  Let  8#=the  price  of  the  horse. 

Then  a — Sx=  chaise.     a=341. 

5 
Per  question,      2« — 16a; — ■3#=24# (a — Sx) 

5 
or,     2a=4Sx (a— 8x) 

14«=301#— 5a+40x 
19a=341#         or,       #=19 

8#=152  Arts. 

(28.)  Let  5x=  his  money. 
After  he  first  lost  and  won  4  $.,  he  had  4#-f4 
He  again  lost  and  won,  and  then  had  3#+3-f-3. 
£  of  this  must  equal  20,   or,     3x-\- 6=24. 

x=6 
5#=30  Ans. 


KEY  TO  ALGEBRA. 

(29.)  Let  3x—  the  income. 
Then    2a?  =  the  family  support. 

x -=-=70.     Hence, ....  3a?=70-9. 

3      o 

30, 31,  32,  and  33  require  no  explanation. 
(35.)  Last  year  the  rent  was  x  dollars. 

_  .  8a; 

This  year  it  is       x-\ =1890 

100 

(36.)  is  the  (35)  in  general  terms. 

(37.)  Let  7a?=  the  income. 

Then    x=  A's  annual  debt. 

7a? 

— =  what  B  saves. 
5 

7a? 


a?=li 

5 

5 

or 

a?=40. 

In  general  terms, 

7a?=280,  Am, 

7a?           a 
~5      X~2 

5a 

x= — 

4 

(38.)                         xx 

3     4 

2a? 

T 

=a 

7a?=£(35d). 

(39.)  Let  10a?=  the  income. 

Then  2a?-f-100=the  sum  spent. 

5a?+  35=  «  sum  left. 
7a?-f-135=10a?  the  whole 

or  45=a? 450  Am 

(40.)  Let  21a?=  the  income. 

Then  3a?-|-«=  the  sum  spent. 

7a?-f&=  "    sum  left. 

10a?+a-K>=21a?=  the  whole. 

(«-H) 

x= . 

11 


EQUATIONS. 
(41.)  2#-h4     :     3x-f-4     : 

(42.)  Let  x* — 7=  the  number. 


Then,  per  conditions,    x — l=Jx* — 7 

a*— 2a?+ 1=^—7 

or,  x=4       and       a?2 — 7=9. 

(43.)  A's  rate  of  travel  is  J  miles  per  hour. 

B's  rate  of  travel  is  f  miles  per  hour. 

A  is  in  advance  when  B  sets  out,  5/  miles. 

Let  a;=  the  hours  after  B  starts. 

5x    7x    56 

Then,     — = — 1 Reduced  gives  .  •  .  •  .  a?=42. 

3       5       5 

CHAPTER  II. 
EQUATIONS  IN  TWO   UNKNOWN  QUANTITIES. 


(6.)  Add  the  two  equations  together,  representing  (x-\-y) 

by  s,  and  we  have     5$-f-3s=50       or 3=5*3. 

But  aH-9y=21-3 

Subtract         x-\-  y=  5*3 

8y=16-3     or, y=6. 

(7.)  By  adding  the  two  equations  we  have 
55=50 
or,  x-\-y= 10 

but  4#-{-2/=34 

Hence,        3a?       =24     or, «a;=8. 

(9)  and  (10)  are  resolved  same  as  (6)  and  (7). 

(11.)  From  the  first  equation  we  have 

y=2x~-  80. 

Transpose  —8  in  the  second  equation  and  we  have 

x-\-v    x    2y — x  , 

9  1  -=— +35. 

5     '  3         4 


10  KEY  TO  ALGEBRA. 

Multiply  by  60  and  we  have 

12aH-122/+20;r=30i/—  15z+35-60 
or          47a?= 183/+35-60 
Substituting  the  value  of  18y,  we  have 
47#=36;r— 1 8-80+35  -60 
or           ll;r=— 240-6+350-6=110-6 
Hence, x=60. 


(14.)  Bringing  unknown  terms  to  the  first  members  of  the 
equation  and  we  have 

4     4_  4     2__3 

x    y  y     x~~2 

2     1 

By  addition,    -=-     or x=4. 

x     2 

(15.)  Put  a=50. 

Then,         #+3a     :     y —  a     :  :     3     :     2 

And  x —  a     :     y-\-2a     :  :     5     :     9 

2£+6a=3y—  3a  (1) 

9x— 9a=5y+10a  (2) 

Multiply  (1)  by  5,  and  (2)  by  3 ;  then, 

10a?+30a=15?/— 15a  (3) 

27#— 27a= 1 5y + 30a  (4) 

Subtract  (3)  from  (4)  and 

17a? — 57a=45a 

17#=102a 

ar=6a=6-50=300. 

(16.)  Divide  the  numerator  of  the  second  member  of 
the  first  equation  by  its  denominator,  and  we  have 

— lla>— 14y+127y 

3*+6y+l=3*+6y+l+        to-J+8 
Hence,  llar  +  142/=127  (1) 

Multiply  the  second  equation  by  (Sy — 4)  and  we  shall 
have 


EQUATIONS.  11 

(151 — 16a?)  (3y— 4) 
9xy—l2x=± '-XZ --f-9ay— 110 

(151— 16#)(3y—4) 

or,  110— 12z=- ili L 

4y— 1 

440t/— 48#y— 1 10+12#=453y— 48#t/— 604-f-64a? 
0=13i/+52;r— 494 
or,  4#+y=38  (2) 

Add  (1)  and  (2),  and  we  have 

l5(x-\-y)=165 
or,  a?+y=ll  (3) 

(3)  from  (2)  gives      3#=27 x=9. 

(17.)  Multiply  the  1st  equation  by  14  and  we  have 
42;r— 7y=49 
Add  — a;+7y=33 

41x        =82     or, «=2. 

(18.)  2 

aH-g2/=16 

5 


2       3 

Subtract  -yA — =19 

3*     5 

Kty+9y=19-15 y=15. 

(19.)  Divide  2d  by  the  1st,  and      x—y~Z 
But  x+y=8. 

(20.)  Multiply  the  first  equation  by  (x+y),  and  the  sec- 
ond by  9,  and  we  have 

4(z-j-y)2=9(za— y*)        9(z*— y*)=9-36 
4(a?-{-y)2==9-36 
Hence,        x+y=9 
Divide  the  1st  equation  by  this  last,  and  we  have 
•  x — #=4. 


12 


KEY  TO  ALGEBRA. 


(21.) 


x=Jy 

3 


a*= 


64i/3 


64y» 
~27~ 


■^=37 


27 
372/8=37-27 y=3. 

(22)  and  (23)  require  no  remark. 

(24.)  The  first  equation  gives 

#4-242/=  91  (i) 

Add         40tf-|-    y=763  (2) 

Multiply  (1)  by  40,  and  subtract  equation  (2),  and 

959y=2877     or, y=3. 

(25.)  From  1st  equation  take  the  2d,  and  we  have 

2hx+5y=60. 
Divide  by  2\  and  we  have        x-\-2y=24 
But  ix+2y=\9 

\x        =  5  or,  a:=10. 
(26.)  Add  the  two  equations,  and 

2(*+y)+£(*+y)+20=*+y 


or,         -+-+20=$ 
2     3 

By  2d  equation, 


5=120. 

£(120)— 5=y=35,  JSm. 


CHAPTER  III. 

EQUATIONS  OF  THREE  OR  MORE  UNKNOWN  QUANTITIES. 


(7.)  Given 


'2x=u-\-y-\-z 

3y=M-f-a?-fz 

4z=u-)rx-\-y 

w=a?— 14 

Subtract  2d  from  the  1st,  and 

2x — Sy=y — x        or 

Subtract  2d  from  the  3d,  and 

4z — 3y=y — z        or 


>  to  find  u,  x,  y  and  z. 


3x=4y        (1) 
5z=4y        (2) 


EQUATIONS. 

Add  3d  and  4th  and  4z=2x-\-y — 14 

Multiply  (3)  by  5  and  (2)  by  4  and 

202=10aH-  5y— 70 
202=  16i/ 


13 


(3) 


0=10a?— lly— 70 

or                      30#— 33i/=210 

But,  equation  (1)     30#— 40iy=     0 

7i/=210 

y=  30 

~x    y    z        * 
2     3     4 

=a+b 

(8.) 

3     4     5 

b 

J>=a — 

4 

x  t  y    z 
.4     5     6 

=a — b 

To  avoid  numeral  multiplication,  and  really,  to  under- 
stand algebra  as  applied  here,  observe  that    62+38=100. 
Put  a=50;  then   62=a-f-12=a+6. 
Clearing  of  fractions,  we  have 

6aH-  4y+  3z=12a-f-12&  (1) 

20#-f- 1 5i/-f-  12^=60a— 156  (2) 

15a?+12y-r-102r=60a— 606  (3) 

Multiply  (1)  by   4  ,  and  subtract  (2). 

Then,    4#+y=636—  12a  (4) 

Subtract  (3)  from  (2,)  and 

5x+3y+2z=  456 


3 


Subtract 


15#-f9i/4-6z=1356 
12a?-j-8y-f-6z=  246+24a 


(5) 


3x+  y         =1116— 24a 
Subtract  (5)  from  (4),  and  we  have 

#=(12a— 486)=12(a— 46)=12-2,  Ms. 


14  KEY  TO  ALGEBRA. 

That  is,         a?=24    or    26. 
Now,  equation  (4)  gives  us 

8b-\-y=G3b—ba 

y=(55— a)6=5-12=60. 
(9.)  By  adding  the  three  equations  and  reducing,  we  have 
4x-\-3y-\-2z=3a  (1) 

By  adding  the  2d  and  3d,  reducing  and  doubling,  we 
have  l0x-\-4yi~2z=4a  (2) 

Subtracting  (1)  from  (2),  and  we  have 

6x-\-y=a  (3) 

Adding  the  1st  and  3d,  and  reducing,  we  have 

x-{-2y—a  (4) 

From  (3)  and  (4)  we  readily  find  x  and  y. 
—2z  =40  (1) 

-\-3z  =35  (2) 

(10.)        <>  3u  +*  =13  (3) 

y  +u  -W=15  (4) 

jx—y-\-3t—  w=49  (5) 

It  is  easier  to  eliminate  t  than  any  other  letter. 
Subtract  (3)  from  (4),  and  we  have 

y— 2u=2  (6) 

Three  times  (3)  taken  from  (5),  gives 

3x—y—\0u=\0  (7) 

Add  (6)  and  (7)  and  divide  the  sum  by  3,  and 

x — 4w=4  (8) 

Double  (6),  and  subtract  it  from  (8,)  and  we  have 

x=2y  (9) 

Eliminate  z  from  equations  (1)  and  (2),  and  we  have 

4#-Hlt/=190 
But4a?=8#.     Then  19t/=190,    or y=10. 


C2x  +y 
4y  — a: 

<  3 


EQUATIONS.  15 


PROBLEMS  PRODUCING  SIMPLE  EQUATIONS,  INVOLVING  TWO 
OR  MORE  UNKNOWN    QUANTITIES. 


(1.)  Let  a?,  y,  and  z  represent  the  num 

a^=600 

a?z=300 

1/2=200 
Multiply  (1)  and  (2),  and  divide  the  product  by 
3,  and  we  have     x2=900 ar=30. 

(2.)  Let  x,  y,  and  z  represent  the  numbers.     Then,  per 
question 


70  (2) 


5 

* +2/+2=l  90  (3) 

Double  (1)  and  subtract  (3),  gives #=50. 

This  problem  calls  the  pupil's  judgment  into  exercise. 
He  does  not  know  in  the  first  place  which  is  greatest,  x  or 
z ;  hence  he  must  try  both  suppositions,  and  the  one  that 
verifies  equation  (2)  is  right. 

(3.)  Let  x,y,  and  z  represent  the  shares,  and  put  a=120 

y — $(x-\-z)=a 
z— §Gr-fy)=a 
Clearing  of  fractions,  we  have 

Ix—Ay — 4z=7a  (1) 

— 3x+8y—3z=8a  (2) 

—2x— 2y-\-9z=9a  (3) 

Double  (1)  and  to  the  product  add  (2,)  and  we  have 

Ha? —  ll*=22a 

X—    z—  2a  (4) 


16  KEY  TO  ALGEBRA. 

Double  (3),  and  to  the  product  add  (1),  and  we  have 
— llH-22z=lla 

— H-  2z=    a  (5) 

Add  (4)  and  (5),  and  we  have z=3a=360. 

(4.)  Resolved  in  the  book. 

Let  x,  y,  u,  and  z  represent  their  ages,  and  s  their  sum. 

Then,  *— *=18 

s—u=ld 

s — y=14 

s — a?=12 


"Rxr  orl/litirm                In        fift  .    ■    . 

,   .  •  t=20 

r>y  auuiuun,         os      — uu  •  •  • 
(«5.)            Let          #=A's  shillings. 

•     •  O^^AiXfrn 

y=B's      " 

z=C's      " 

After  the  first  game  they  will  have  as 

here  i 

•epresented 

x — y — z=A 

2y            =B 

2z            =C 

After  the  second  game, 

2x— 2y— 2z-—k 

Sy —  x —  z=B 

4z                =C 

After  the  third  game 

4a? — 4y — 4z=  16 

CD 

6y—2x—2z=  16 

(2) 

Iz —  x —  y=  16 

(3) 
(*) 

Sum,            x-\-  y-\-  2r=3'16 

Add  (3)  and  (4)  and  we  have 

8z=4  16 

.  .  .  z=8. 

(6.)  This  problem  is  resolved  in  the  book,  by  equation 
7,  Art.  53. 


EQUATIONS.  17 

(7.)  Let  x  represent  the  better  horse,  and  y  the  poorer. 

ar+10=H(y+ls) 
Therefore,        J(y+10H|(y+15)+5 
Reduced  gives i/==50. 

(8.)  Let  #=  the  price  of  the  sherry. 

y= brandy. 

Put        a=78. 

2a?-}-  y=3a 

7x+2y=9a+9 


3a?         =3a-f-9 

x        =  a+3=.  .  81a 

(».)  Let  x=X's  time.        y=B's  time. 

Then, 

1                                                     , 
-=the  part  that  A  can  do  in  one  day. 

X 

And 

1 

_=the  part  that  B  can  do  in  one  day. 

y 

4     4      4      1 
x    y     16~~4 

36     3 

— =-     Hence 

y    4 

(10.) 

2a?       2                          x-\-2     3 

y+7     3                           2y       5 
3a?=y-f-7                      5a?+10=6y. 

(11.) 

,  2y                           ,  3* 

(12.) : 

Let  x=  the  greater,  and  (24 — a?)=  less. 

a?              24 — x 

..4.1 

24 — a?             a? 

a?2     :     (24— a:)2     :  :     4     :     1 

By  evolution,  x     :     24 — x     :  :     2     :     1. 

♦   2 

y=48. 


18  KEY  TO  ALGEBRA. 

(13.)  Let  a?=  the  number  of  persons. 

yz=  what  each  had  to  pay. 
Then,     xy=  the  amount  of  the  bill. 

Put     (x+4)(y—  1)= the  bill. 

Also,  (as— 3)(y+l)= the  bill. 

xy + 4i/ — a? — 4 = #?/ 

4y — x — 4=0 
— 3y-{-x— 3=0 

By  addition,  y      — 7=0 

(14.)     \0x-\-y=4x-\-4y     or, y=2x. 

\0x+y+27=l0y-\-x 
or,  9x       +27=  9y 

x       +  3=     y=2x  hence,  a?=3. 
(15.)  Let  rr=  the  digits  in  the  place  of  100's. 
y=            "      in  the  place  of     10's. 
z=           «*                 the  units. 
x-\-y+2=ll                                z=2x 
100.r+10y+2r+297=100z+10i/+a: 
99^+297  =  99* 
x-\-3=z=2x       Hence x=3. 


x— -40       #—20  ,     #—10 

(16.)  Let     ,      ,     and represent 

2  3  4 

the  parts.     Then 

x — 40     x — 20     a?— 10 

1 1 =90 #=100. 

2  3  4 

(17.)  Let  x  represent  the  part  at  5  per  cent,  and  (a — x) 

the  part  at  4  per  cent.     Then 

bx     4a— 4x     . 

+ =6 

100         100 

Hence x=100&— 4a. 


EQUATIONS.  10 

(18.)  To  avoid  high  numerals,  and  of  course  a  tedious 

operation,    Put  a=5000 ;    then   2a=10000,  3«=15000, 

3a  16a 

— =1500,  and =800. 

10  100 

Put  #=  A's  capital,  and  r — 1  =  A's  rate. 

x-\-2a—  B's      "  r=  B's  rate. 

a?-j-3a=  C's      "  r+l=  C's  rate. 

~rx — x     16a     rx-\-2ar 


By  conditions, 


100       100         100 
rx — x     3a     rx-\-3ar-\-x-^3a 


100        10     .  100 

Reducing  (1),  gives  a?=(16 — 2r)a 
'27— 3r 


(1) 

(2) 


/27 — 3r\ 

"    (2),  "  *=(— — y 

Hence,     32 — 4r=27 — 3r,       or r=*=5. 

(19.)  Put  a=1000,  x  and  y  to  represent  the  tw<  parts, 
and  r  and  t  the  rates  expressed  in  decimals, 

Cx+y=13a (i) 

Then  by  conditions,^                  y '                              "  ^  ' 
'  ]        te=360 (3) 

t     ry=490 (4) 

Divide  (3)  by  (4),  and  we  have 

36 

(5) 


C-)QJ 


49 

x    t 
From  (2)  we  have         -=- 

y    r 

Substitute  the  value  of  -  in  equation  (5),  and 

T 

x*     36  6v 

--= —  or  x= — 

y*     49  7 

6v 
By  returning  to  equation  (1)  we  have    -— +2/=13e*. 

•13y=13a*7      or. y=7a. 


20  KEY  TO  ALGEBRA. 

(20.)  Let  x,  y,  and  z,  represent  their  respective  ages. 
Then  by  conditions  given,      x — y=     z 
5y+2z—  x=U7 
x+y+z=  96 
(21.)  Let  x,  y,  and  z,  represent  the  respective  property 
of  each,  and  put  s=their  sum. 

rx+3y+3z=47a  a=100. 

Conditions,  J  y_|_ 4X+4z=58a 
Lz+5x-\-5y=63a 
Add  2x  to  the  1st  equation,  3y  to  the  2d,  and  iz  to  the 
3d,  observing  that  x-\-y-\-z=s  ;  then  we  shall  have 

3s=47a-f-2;z (1) 

4s=58a+3y (2) 

5s=63a-f4z  . (3) 

3s— 47a 


or, 


2 
4s — 58a 

5s — 63a 

z= 

4 

3s — 47a     4s— 58a     5s— 63a 

By  addition,  s= h H 

\  2  3  4 

Hence, s=19a. 

This  value  of  s,  put  in  equation  (1),  gives  #=5a=500. 

(22.)  Taken  from  the  book  we  have 

Is — p     ms — a     ns — r 

s=-—f-+ 1+ 

/ — 1      m — 1      n — 1 

or,  ,~(±.y.r+(JL Y__i_ +(JL y__ L. 

V/-1/     /_i     \ot-1/      m-1     \nr-l/     n-l 

(I          m           n          \          p          q            r 
— + -f- 1  )s=-i--r— H—  -f 


EQUATIONS.  21 


4-1-  ?■■!■■.»- 

/ — 1     m — 1     w — 1 

Hence,     s= =k 

I  m  n 

I — 1     m — 1     n — 1 

As  s  is  now  known,  we  call  its  value  k. 

lk—P 
Then, #= -- 

mk — q 

nk — r 
r— 1 
(23.)  Let  x,  y,  and  z  represent  the  respective  sums. 

x+ra (i) 

y+5=« (2) 

*+i=« (3) 

2x-\-y=2a 

3y-\-z=3a        or,        4z-\-\2y—\2a 

4z-\-x=4a        or,        4z±     x=  4« 

— x+l2y=  8a 
From  the  1st  24x-\-l2y=24a 

25#  =16a 

(24.)  This  problem  is  resolved  in  the  work,  by  the  13th 
example,  page  80,  (Art  51.) 

(25.)  Let  x=  the  greater,  and  y  the  less. 
5#-Hy=13 
jx — 5y=0     or, 2x=3y. 


22  KEY  TO  ALGEBRA. 

(26.)  #-K(y+z)=a=51 

y+*{x-\-z)=a 
z-\-k(x+y)=a 
x+(x-\-y+z)=2a 
2y+{x+y+z)=3a 
3z-}- (x-\-y-\-z) =4a 

x—2a — s (1) 

y=i(3a— s) (2) 

*=i(4a— 8) (3) 

s=  2a —  s-f-£(3a — s)-H(a — s) 
6«= 12a— 6s+9a-— Ss-\-Sa—2s 
I7s=29a    or 5=29-3=87. 

Now  equations  (1),  (2),  and  (3),  will  readily  give  x,  y, 
and  z. 

(27.)  Let  #=A's,  y  B's,  and  z=C's  sheep. 
Then  by  the  conditions, 

x+8 — 4=y-fz— 8 

5(y+8)="ar+ar— 8 

*(*+8)=ar+y— 8 

*4-12=»  y+* (1) 

y+24=2x+2z (2) 

z-\-32=3x-\-3y (3) 

Add  (1)  and  (3),  and  we  have    x-\-44—3x-\-4y. 
Double  (1),  and  subtract  (2),  and  we  have 

2x — y—2y — 2a?  or,  4ar=3y 

But 2#-f4y=44 

4#+8i/=44-2 

lly=44-2     or y=8. 

(28.)  ;r+l_l  x    _l 

"J        3  y+l~4 

(29.)  a?+2__5  a:    _1 

"^        7  ^"S 


EQUATIONS.  S3 

(30.)  This  is  a  repetition  of  the  10th  example,  page  89, 
inserted  here  by  oversight. 

(31.)     Let  #=A's  money,  and  2/=B's. 

*-5=i(y+5) (1) 

ar-r-5=3y — 15 (2) 

Subtract  (1)  from  (2),  and  we  have 

10=3y— 15— or y=ll. 

2     Z 

(32.)  Let  a?=  the  number  of  bushels  of  wheat  flour. 

And        y= barley     " 

Then  the  cost  of  the  whole  will  be  expressed  by 

iOx-\-4y 
The  sale  at  11  shillings  will  be     llic-f-lly 
Now  by  the  conditions, 

10aH-4y    :     ll#-f-lty     :  :     100  :  143| 
Multiply  the  last  two  terms  by  4,  and 

\0x-\-4y     :     llx-\-lly     :  :     400  :  575 
Divide  the  two  last  terms  by  25,  and 

10aH-4y     :     ll^-flly     :  :      16  :  23 

bx+2y     :     llz-fllt/     :  :        8  :  23 

115#-}-46y=88;r-f88i/ 
27^=42y 
9x=-\4y 
These  co-efficients,  9  and  14,  give  the  lowest  proportion 
in  whole  numbers.     The  proportion  was  only  required. 
(33.)  Let  lOx -jry  represent  the  number. 

Kito+sO-e-H 

i(io*-h/)=<M-i 

Now  the  question  gives  us    Q  =2x 

And Q'^ty 

.        i(10*-f-y)=2ar-H    or y=l. 


24  KEY  TO   ALGEBRA. 

INTERPRETATION  OF  NEGATIVE  VALUES. 

(Art.  55.) 

(4.)  Let  x  represent  the  years  to  elapse. 

Then     30-r-ar=3(15+a?) x=— lh 

To  make  this  equation  true,  the  years  required  must  be 
taken  subtr actively. 

(5.)  Let  x=  the  man's  daily  wages,  and  y—  the  son's. 

7x+3y=22 (1) 

5#-f-  */=18 (2) 

12aH-4?/=40 
Sx-{-  y—lS        #=4,    y= — 2. 

(6.)  Let  x—  man's  wages,  y=  wifes,  and  z~  the  son'? 

10#+  8y-\-  6^  =  1030  cts (1) 

\2x-\-l0y-{-  4z=1320cts (2) 

\5x+l0y-{-  12sr=1385cts (3) 

Subtract  (2)  from  (3),  and  we  have 

3x+8z=65 (4) 

Multiply  (1)  by  (5),  and  (2)  by  4,  and  take  their  differ- 
ence, and  we  have       2x-\-\4z=      130 

x-{-  7z=—    65 (5) 

Sx+21z=— 3-65 
(4) Sx-\-  Sz=        65 

By  subtraction,  13z= — 4*65 

z=:_4.5=— 20 
As  z  comes  out  with  a  minus  value,  it  shows  that  the 
eon  had  no  wages,  but  the  reverse  of  it,  he  was  on  expense. 

(7.)  10;H-4y-r-3z=1150 (l) 

9£-r-82/-f  6z=1200 (2) 

7x+6y-{-4z=  900 (3) 

Double  (1),  and  subtract  (2),  and 

11^=1100 a?=100cts. 


EQUATIONS.  25 

(8.)  x+ l_3  x    __5 

~~^     5  y+f" 

5X-\-5=3y (1) 

7x       =5y+5 (2) 

Add 12a;       =8y 

3^       =2y (3) 

Subtract  (2)  from  (1)  and  we  have 

—2^+5=— 27—5 (4) 

#=__10  by  adding  (3)  and  (4). 

y=— 15 
The  result  coming  out  minus,  shows  that  there  is  m 
such  arithmetical  fraction.     Algebraically,  however,  »~tj 
will  answer  the  conditions. 

SECTION      III. 

We  pass  over  the  first  three  chapters  of  this  section,  as 
requiring  no  aid  from  a  key. 

We  commence  at  the  examples  in  approximate  cube  root 
*t  page  126. 

(1.)  What  is  the  approximate  cube  root  of  120  1 

125  has  a  root  of  5. 
2 

250  240 

120  125 

370     :     365     :  :     5     :     root  required. 

(2.)  What  is  the  approximate  cube  root  of  8.5  ? 
8  has  a  cube  root  of  2. 

16  17 

8.5  8 

24.5     :     25     :  :     2     :     root  required. 


26  KEY  TO  ALGEBRA. 

(3.)  What  is  the  approximate  cube  root  of  63  ? 
64  has  a  cube  root  of  4. 
128  126 

63  64 

191  :  190  :  :  4  :  root  required, 
(4.)  What  is  the  approximate  cube  root  of  515? 
512  has  a  cube  root  of  8. 

1024  1030 

515  512 


or 

1539     :     154 
1539     :        3 

sur: 
J980*  = 

3j54xA  = 
43>/708  = 

\2     :  :     8     :     cube  required. 
:  :     8     :     correction. 

DS  IN    GENERAL. 

(2.) 

:  ^/49a2 

X2 

=  7aj2 

(3.) 

=  JWXZy 

=2xjSy 

(4.) 

:y27£3: 

K2x 

=3x*j2x 

(*•) 

=4  727. 

X4 

=  12  V4 

(6.) 

Jx? — aV= 

3JS2a3    = 

J2$a3x2= 

=  Jx2(x- 

-a*) 

—xjx — a* 

(*•) 

=  V803 

X4  i 

=2a  V4 

(8.) 

=  tj4a2x2 

X7a 

=2aav7a 

(9.)  Performed  in  the  work. 

V    64        V    64        V   64  4V 


(12.)  /50  /25-2  /25-6  5     _ 


PURE   EQUATIONS.  $7 

(i3.)  *  v^^6"2=v^(TT6T=«  vh^ 

(14.)  /2a         /T  1    — 

3  9  3V 

(Art.  81.)  Requires  no  aid  from  a  key. 
(Art.  82.) 

(I.)         5^5X3/8  =  15^X2X4=30710 
(2.)         4712x3^2-12724=24^/6 
(3.)         372  X  278=6716=24 
(*•)         2  yuX3  374=6377X8=12877 

(5.)      275  x  27X0=4750=2072 

(1.)  Here  to  multiply  (a-f-^)3   by  (a-f-6)    we  must  sim- 
ply add  the  exponents. 

(2.)         (7)^X7^=7^^=7^ 

(3.)      2(3)*  X3(4)*=2(3)^X3(4)*=2(27)*X  3(16)*== 
6(27X16)*=6(432)~« 

(4.)     (15)*X(10)*=(15)*X(10)*=(225)*X(1000)*= 
(225000)* 

PURE  EQUATIONS. 

(14.)  Multiply  both  members  by  Ja-\-x%  and 
tJax-Jrx2-\-a-{-x=i=2a 
jjax+x^a — x 
ax-\-x*=a2 — 2axJrx% 
3ax=a* x*=la. 


KEY  TO  ALGEBRA. 


(15.)  Multiply  by  Jtf+x2,  then 


xJa2+x2Jra2-\-x2=2ai 
xJa2-\-x2=a2 — x2 
By  squaring  both  members  and  reducing  we  have 
Sa2x2=a4    or x=±ajl, 

(16.)  Square  both  members,  and 

x2+2ax+a2=a2+xj¥+x* 
Drop  a2  and  divide  by  x,  and 

x-\-2a=Jb2-\-x2         Square,  &c. 

(17.)  Divide  the  numerator  by  its  denominator,  in  each 

member,  and  we  have 

4                       15 
1, =1 


v/6a?-r-2           4vQx-\-Q 
Drop  1  and  change  signs,  and  clear  of  fractions,  and 
16^6^-1-24=1576^+30     Hence x=6. 

(18.)  Cube  both  members,  and 

(4-\-xf 

64-f-z2— Sx=- -=(4+x  )2 

(4+*)      K  T    J 
Hence,  64-\-x2—8x=16-\-8x-\-x2    or x=3, 

(19.)  By  clearing  of  fractions,      f>-\-x-\-  J  x2-\-5x=\5 
By  reduction,     1Jx2-{-5x=lO — x     Square  &c. 

A/  x-f-Jx — \/  x — Jx= -—= 


V x+Jx 

M  iltiply  by  \f  x-\-  Jx,  and  we  have 



X   \    w  X        i^  Ju         *L        2  /^  X 

2a:+ 2  Jx — 2  Jx2—x =3jx 

2x — Jx==2s/x2 — x 

4X2 — 4x,Jx-{-x=4x2 — 4a?  *=H* 


PURE  EQUATIONS.  29 

(21.)  Resolved  in  the  work. 

(22.)  Resolved  the  same  as  17. 

2b                      7b 
1 =1 


a 


\/ax-\-b  3Jax*\-5b 

_                    _  96* 

Hence,     6Jax+l0b=7jax-{-7b  or x- 

(  U.  136.  )  (  &  136.  ) 

(23.)  Square  both  members  and  we  have 
^+a^2-f-12=/i+2;r+z* 
lJx2-\-\2=2-\-x  Square,  &c.  .    x=2. 

(24.)  Multiply  numerator  and  denominator  by  the  nu- 
merator, and  we  have 

i 

Take  square  root,  and  transpose  ,j4x,  and 
N/17-Fl=3— J4x 
Square,         4a?+l=9 — 6,j4x-{-4x    Hence,  #=£. 

(25.)  Square  root  gives    a — x=Jb    or,  x=a — Jb. 
(26.)  Clearing  of  fractions  and  we  have 

271—^=73 

4— 4;z2=3  x2=% 

(27.)  Take  the  square  root  of  both  members,  and 

2        1 
=-     or 4=a? — 1 

x—l     2 

(28.)  Resolved  the  same  as  the  21st. 
(29.)  Clearing  of  fractions  we  have 
A/ri^M-a>-9=36 


30  KEY  TO  ALGEBRA 

Jx2 — 9#=45 — x 

x2 — 9^=452—90^+^ 

81z=45-45 

9-9#=5-9-5-9 #=5-5=25. 

(30.)  Resolved  the  same  as  (17)  and  (22.) 

Dividing  numerators  by  denominators,  we  have 

10                  10-5 
3 =3 


jx+2        V^-Ho 

Drop  3  from  both  sides,  change  signs,  and  divide  by  &, 
and  clear  of  fractions,  and 

2>/#-f80=21>/x+42     Hence x=4. 

(31.)  Multiply  numerator  and  denominator  of  the  first 
member  by  (Jx+Jx — a) 


Then,        y+</~* 


a  x — a 

Multiply  by  «,  and  take  square  root,  and 

an 


Jx+  Jx—a= 

Jx — a 

Jx2~ax-\-x — a=an 
Jx2 — ax— (n+ 1  )a — x 
xi—ax={n+\)2a2—2a(n+\)x+x% 
Drop  x2,  and  divide  by  a,  and 

— x = (n  -f- 1  )2a — 2nx — 2a? 
(l+2n)x=(n-t-l)*a 
_(n+l)2« 

X  l+2ra 
(32.)  Resolved  in  the  work. 
(Art.  90.) 

(4.)  Observe  180=9*20     189=921.     Put  a=9. 
*fy+xy*=2Qa  sHy=21a 


PURE  EQUATIONS.  31 

Multiply  the  first  equation  by  3,  and  add  it  to  the  second, 

and     x3+3x2y+3xy2-\-y;i=81a=a!i 

cube  root,  a?-f-y=a=9 

The  rest  of  the  operation  is  obvious 

(5.)  Divide  the  first  equation  by  (x+y)  and 
x2—xy+y2=xy 
or    x* — 2xy-\-y2=0  or    x — y=0. 

Hence x=2    y=2 

(6.)     x+y     :     x     :  :     7     :     5  xy+y2=l2& 

5x-\-by—7x 

5y=2x        or     a?=§i/. 
Put  this  value  of  x  in  the  second  equation,  and 

7y2=  126-2 
^=18-2=36   .  .  .y=±6. 
(7.)  From  the  first  equation  we  have 

5x—  5y=fy  5x—9y 

xz-{-4y2=lSl 
1^+4^=181 
8h/2-r-100t/2=181-25 

181#2=181-25   or i/2=25. 

(8.)  From  the  proportion  we  have 

Jx+Jy^Jx— ±jy 

§Jy—%jx    or 25#=9#. 

The  rest  of  the  operation  is  obvious. 

(9.)  Extract  square  root  and 

|jp-H=3     or • x=-lh 

(10.)  From  the  first  proportion 

x-\-y=Zx — dy  or   4y=2x 

Hence 8y>=x*. 

8V3— 1/3=56  y=8  y=2. 


32  KEY  TO  ALGEBRA. 

(11)  (12)  and  (13)  resolved  in  the  work. 

U4.)      or2— v8 

-=6    or a?+v=6 

x—y  * 

From  the  second, #y=5. 

(15.)  Divide  the  1st  equation  by  (x-\-y),  and 
x2 —  xy-\-yt—2xy 

x2 — 2xy-\-y2=xy=lQ (1) 

Add  4xy  =64 

a?-\-2xy+y*=         80=16-5 
Square  root,  x-{-y=4)j5 

Square  root  of  (1)   x — y=4 

2x       =4^5+4 
(19.)  Double  the  2d  equation,  and  add  and  subtract  it 
from  the  1st,  then 

x2-\-2xy-\-y2=a-\-2b 
x2 — 2xy-{-y2—a — 26 
z+y=Ja+2b 
x—y=Ja — 2b 

(Art.  92.) 

(5.)  Add  the  two  equations  and  extract  square  root,  and 
i       i 
we  have  x* -\-y*  =±4 (1) 

Separate  the  first  member  of  the  first  equation  into  fac- 
tors, and  we  have  #5Cz5+i/5)==12 (2) 

Divide  (2)  by  (1)  and    a?*=d=3 x=9. 

(6.)  is  of  the  same  form  and  resolved  the  same  as  (5.) 
(?.)  Add  the  two  equations,  and  extract  square  root,  and 

3  3 

we  have  x* -\-y*  =  J a-\-b 


\ 


PURE    EQUATIONS.  33 

But a?*  Gr* +#*)=« 


I          a 
ar  = 


Ja+b 

x?— or x=  ( j 

(a+bf  \  («+6)V 

(8.)  Resolved  in  (Art.  90,)  of  the  Key. 
(9.)  Square  the  first  equation,  and 

x  -fy^ia (1) 

Difference 2x*y*       =  12 (2) 

Subtract  (2)  from  (1)  and 

x— 2x*y*+y  =  1 

By  evolution x* — y*  =±1 

But x*+y*  =     5 


2a?5  =6  or  4. 

CHAPTER    V. 
(Art.  93.) 

QUESTIONS  PRODUCING  PURE  EQUATIONS. 

(5.)       Let      x-\-y  =  the  greater  number, 
And  ......   x — t/=  the  less 

Difference  •  .  .  .2^=4  Sum  =2x 

2a?(4#i/)  =  1600     Hence.  .  .  .#=10. 

(6.)  Let  x-\-y=  the  greater  number. 
x — y=  the  less. 
2y     :     x-—y     :  :     4     :     3     or x= \y. 

(     x2— y2)(  x—y)=504 

(W)  (}y)  =504     Hence  y=4. 

3 


34  KEY  TO  ALGEBRA. 

(7»)  Let  Sx=  the  length  of  the  field,  and  5x=  its  breadth 

40^     x2 

Then =— =  the  acres. 

160      4 

?#2X8;r=the  whole  cost. 

26x—  the  rods  around  the  field. 

13X26#=the  whole  cost. 

Hence  2xz=lS-26x    or    #=13    .  .  .  8x=104,rfns. 

(8.)  Let  5a?—  the  length  of  the  stack. 

4x=  the  breadth. 

7x 
Then,       — =  the  height. 

2 

Ix 
5x'4x* — -4x=  the  cost  in  cents. 
2 

Also,       5;r-4av224=  the  cost  in  cents. 

Ix 

Hence,     5x*4x' — •4#=5a:,4a:,224 

2 

7a?-2ar=224     or x—4. 

(9.)  Put  x2 — 7=  the  number. 
Then        x-\-JxJ+9^=9 

^2-j_9— 9— x    or x—4, 

(10.)  Let  x  represent  A's  eggs ;    then  100 — x—   B's 

18  8 

=  A's  price.  -=B's  price. 


100— x 


18#        8 1 

Hence =-(100— #) 

100— x     xx 

9^2=4(100— xf 

3#=2(100— x) a?=40. 

(11.)  Let  x-\-y=  the  greater  number, 
x — y=  the  less. 

2#=6      or     x=S 


QUADRATIC  EQUATIONS.  35 

(x-\-yY=x3+3xzy+  3xy2+y* 
(x^yf=xa-^3x2y-{-3xy2—yt 

2x3  +6xy2=72 

Divide  by  2x  and     a*+3y2=12     Hence  .  .      »y=l. 

(12.)  Let  x=  one  number 
Then      a2x=  the  other. 


b 

aV=62  arr=6  x=-. 

a 

(13.)  Let  a?3  and  y2  represent  the  numbers. 

Then a?2+i/2=l00 

x  -\-y  =  14 

SECTION   IV. 

QUADRATIC  EQUATIONS. 

None  of  the  examples  require  the  aid  of  a  key  until  we 
come  to  the  12th,  Art.  101. 


(12.)  Put  x=\u.     Then  5x2= jW2,  and  the  equation  be 

comes \u2-\-\u=  273 

m2+4m=1365 
Hence  w=35  or  — 39  which  gives  .  .  .  x=7  or  — 3j°. 

(13.)  Put  x=\u.     Then 

u2 — 20w  =224 

W2__20u-f-100=324 
w_10=±18 

«=28  or  — 8     and  •  •  x=7  or  — } 
(14.)     25#2—  20#=— 3. 
Complete  the  square  by  (Art.  99.) 

25a;2— 20a?+P=/2—  3 
5atf=— 10a?             t=—2             t2=4 
5x—2-=^zl x=i  or  \. 


36  KEY  TO  ALGEBRA. 

(15.)  Put  x=^u.     Then 

«2 — 292u= — 500-21=— 10500 

w2— 292n4-(146)2=108L6 

w— -146=±104 

w=250  or  42 

Hence »=Vt°  or  2. 

(Art.  103.) 

EXAMPLES. 

(1.)  Put(a?+12)*=y.     Then 

2/2+2/=6 2/=2  or  —3. 

(a?+12)*=2or  —  3 

a:-|-12  =  16or81 a?=4  or  69. 

(2.)  Add  b2  to  both  members  and  extract  square  root 
re  then  have         (rc-f-«)*+&=rt2& 
The  rest  of  the  operation  is  obvious. 

(3.)  Put  y=(9x+4f     Then  tf+2y=\b 
y=3  or  — 5     Hence 9aH-4=9  or  25. 

(4.)  Put    y=(l0-\-xf. 

Then  y2— y=2 y=2  or — 1 

(Art.  104.) 

3 

(5.)  Put.  .  .  .(x — 5)2=?/;  then 
y* — 3^—40 y=S  or  — 5. 


(6.)  Put     (1+37— x*)*=y;  then 

.2 ,,— : 1 


2tf 

From  which  we  obtain  y  equal  5  or  \. 
Hence  .  .  l-\-x — afej-  or  ^.     From  which  we  obtain 
x=4t±$jU  or  izt&.fm. 

(7.)  Put.  .  .  .  (a?+ 16)^=1/ 

^2 — 3y=io     Hence   y=5  or  — 2. 


QUADRATIC    EQUATIONS.  37 

(§.)  Puty=a?n;  then 

d*f—2y=B  Put        y=\u 

w2 — 2w=24  w=6  or  — 4 

y=Z     x=nJ2. 
(9.)  Multiply  by  4,  &c.     Rule  2. 

4a?*+4a?M- 1=3025 

2a?*+l=±55 

,                   a?*=27  or  —28 
a?s=3    or a?=243. 

(10.)  Put  (2a?— 4)2=y. 

8           16 
Then -=H 

3/      y2 

8y=^»+16 
^— .8i/+16=0     or y— 4=0. 

(11.)  Multiply  by  16.     Rule  2. 

Then  .   .  .  64a?^+16a?*+ 1=39- 16+ 1=625 

8a?i+l=25  a?*=3  ;r=729. 

(12.)  Add  5  to  both  members. 

Then  {x2— 2H-5)+6(ar2— 2a?+5)*=16 

By  subtraction,    2/2+6y+9=25 y=2  or  — 8. 

Hence   ar2 — 2a?+5=4 a?=l. 

(13.)  By  (Art.  99)  we  have 

x*      12a? 
1_,2=/2_32 

361      19 
a?  6a? 

— 1= *=— 6  *2=36. 

19  19 

x2      12a? 

1-36=4 

361      19 

a? 


19 


-6=±2    a?=152  or  76. 


38  KEY  TO  ALGEBRA. 

(14.)  Observe  that  81 2*  and  —  are  both  squares,  and 

ar 

if  these  are  taken  for  the  first  and  last  terms  of  a  binomial 

1 

square,  the  middle  term  must  be  9a?*-'2^18. 

x 

This  indicates  to  add  1*  to  both  members.    Then  extract 

1 
square  root    9x-\ — =±10     Hence.  •  •  •  x=l  or — 1. 
x 

(15.)  The  first  member  of  (15)  is  the  same  as  (14.) 

Hence,  add  unity  to  both  members  and  extract  square 

1     29 

root ;  we  then  have      0x-\ — = j-4 

x     x 

u 
932— 43=28  Put  3=- 


9 


w2— 4w=28-9=252 

u — 2=±16 3=2. 

(Art.  105.) 

(4.)  Multiply  every  term  by  3,  and 

^_833+19a:2— 123=0 
Operate  for  square  root  thus 

a?4— 82?+ 19a*— 12a;  (a* — ix 


2x*— 4x)— 82?+  19a?2 
— 833+1632 


332— 123 
3(a*— 4x) 

(a?— 43)2+3(32— 43)=0 
Divide  by  (2* — 4a:)  and 

2*— 43+3=0 3=1  or  3 

But  the  factor  2* — 43  gives  3=0  or  4. 


QUADRATIC  EQUATIONS.  39 

(5.)      #4—  10x3+35a^--50^4-24=0  (  x2—  5x 


2x?— 5x)— 10x3+35x2  jr&     OY  THE 


UNlVEESITY 


10x2— 50*+24  v    /y 

(a^— 5ar)8+10(a?»— 5a?)+24^^XrIpQB,,»VJ 
If  we  add  unity  to  both  members,  we  shall 
plete  squares.     Extract  square  root,  and 

x*—5x= — 4  or  —6 
From  these  two  equations  we  find  #=1,  2,  3,  or  4. 

(6.)  By  mere  inspection  we  perceive  that  this  equation 
can  take  the  form     (a:2—- a?)2— (a?2— #)=132. 

y2—y=l32 #=12  or — 11. 

x2 — x—Vl  or  — 11 
If  we  take  — 11,  the  value  of  x  will  become  imaginary, 
12  gives  #=4  or  — 3. 

(  U.  169.  ) 
(7.)  This  equation  may  be  put  into  this  form  : 
(y»_ cy)-—  2(y2—cy)=c2 
from  which  the  reduction  is  easy. 

(Art  107.) 

(3.)  Taken  from  the  work  we  have 
(a+  l)x2—a2x=a* 

or (a-\-l)x2=(x+l)a2 

Both   members  are  of  exactly  the  same  form,  and  of 
course  the  equation  could  not  be  verified  unless  •  .  a?=«. 

EXAMPLES. 

(1.)         #2  +  llx=80.     Multiply  by  4,  &c. 
4.z2-f-./H-ll2=320+l21=441 
2#-f-ll=±21 x=5  or  —16. 


V\  KEY  TO  ALGEBRA. 

(2.)  Drop  2x  from  both  members,  and  divide  by  3  ;  then 

x — 1     x — 2 

x = 

#— 3        2 

Clearing  of  fractions  and 

2X2— 6;r— 2x-\-2=x*— 3x—2x-\-6 
x2—  3*=4  Put    2a=3 

Hence,  (Art  106) x=4  or I 

(3.)  Multiply  the  equation  by  6x  ;  then 

6x2 

— ■ |-6tf-j-6=13a? 

x-\- 1 

Gx2 

_4-6=7* 
a:+l 

6x2+6aH-6=7;r2-r-7a: 
Hence  x2-^-x—6 x—2  or  — 3. 

(4.)  Clearing  of  fractions 

70#— 2  \x2  +72^=500— 1 50a? 
21#2— 292*=— 500 


(5.)  Put    f-+y\=x.     Then 

a?3-f-a?=30    or, x=5  or  — 6. 

6 
Now, — H/=5  or  — 6 

y 

y2—  5y=— 6     or     t/2+6?/=— 6 

Ay2—-  .#+25=25— 24 

2y— 5==hl y=3  or  2. 

(6.)     Put  x%=y;    then    i/2+7#=44 
Ay2  +.#+49=225 
2y+7=±15 2/=4or— 11. 

a?=(4)f  or  (— 11)*. 


QUADRATIC   EQUATIONS.  41 

(7.)     x2-{-x=42.     Hence  x=6  or  — 7. 

That  is  yH-1 1=36     49     or y=5  or  «/38. 

(8.)  ar+7     a: 

x—1     3 

33a? — 23 1— 3a?— 2 1  =x*—-lx 

x2— 37a?=—  252 

4a?2— .tf-I-372=1369— 1008=361 

2a;— 37==t=19 *=28  or  9. 

(9.)  3a?2— 9a?=84 

12 


3622— ^+81  =  12-84-1-81  =  1089 
Gx— 9=±33. 

(10.)  Clearing  of  fractions  we  have 

2x+2tJx=\6—x 

^x-\-2s/x=\G 
Multiply  by  12,  &c. 

6^/a?-{-2=±14 a?=2or7j. 

(II.)         6(2a:—  li; 


a? — 3 

3(2a?— 11) 


•4a?=26 


-2a?=13 


a:— 3 

6a?— 33-f-2a?2— 6a?=13a?— 39 
2a?2— 13a?=— 6 
16a?2— ^-1-132=169 — 48=121 
4x— 13=±11. 

(12.)  Multiply  by  x*  and  we  have 

22a:2 

10a: — \4-\-2x— 

9 
11a:2 

6a:— 7= 

9 

11a?2— 54a: =—63 


42  KEY  TO  ALGEBRA. 

u 

Put   x=— ;  then    u2 — 54m  =—693 
11 

u2—542i-\-212=3Q  s 

u— 27=±6 m=33  or  21. 

(13.)  Clearing  of  fractions  we  have 

£*_!  o#2-f- 1=*3— 6^2~|-9^— 3x2+ 1 8z— 27 
— #2=27.r— 28 
tf2-f-27x=28         Put  2a=27 
a'2-f2a^=2«+l 
x2+2ax-{-a2=a2-{-2a+l 
a?-fa=±(«+l)    •  .a?=lor—  28 

(14.)  Given  mx2 — 2mxjn=nx2 — rnn  to  find  x 

By  transposition  mx2 — 2mxJn-\-mn=nx2 
Square  root  .  .  Jmx — Jmn—dcjnx 
By  transposition  {Jm±Jn)x=,Jmn 

Jmn 


CHAPTER  II. 

QUADRATIC    EQUATIONS    CONTAINING    TWO    OR    MORI 
UNKNOWN     QUANTITIES. 

Problems  1,  2,  and  3,  require  no  aid  from  a  key 
(4.)  Put  x=vy;  then  the  equations  become 

v*y*+y*  =  18vy* (1) 

vy+y=\2 '  •    .    .    .  (2) 


QUADRATIC   EQUATIONS.  43 

From  (1)  we  have y— 

*    «M-i 
12 

From  (2)  we  have y= 

v-f-1 

Hence = 

v-f-1     v3-f-l 

3v 
Divide  the  denominators  by  (v-f-1),  and     2=—-  • 

Or,     2v2 — 5v= — 2 v=2  or  |. 

•Another    Solution. 
Cube  the  2d  equation  and  we  have 

x3-\-ij3-\-3xy(x+y)  =  l2s 
That  is.  .  .  .  18x'i/H-3xi/(12)  =  12-12-12 
Divide  by  6,  and  we  have  3xy-\-6xy=2'\2-l2 

Or 9xy=2'3'4-34 

Or     ^=2-4-4=32 

Now,  having  (x-\-y)  and  (xy),  the  rest  is  obvious 

(5.)  The  first  equation  can  be  put  in  this  form 

(x+y)2+2(x-{-y)  =  l20 
Or s2-f2s-f-l  =  121 

*+l=dbll 

x-\-y=\0  or — 12 
Or a?=10 — y   or   x= — y — 12 

From  the  second  equation       x= — 

y 

8+y2                             8+v2 
Hence  .  . _=10— y     or,     — = — y J  2 

y  y 

8-|-2i/2=:10i/     or,     8+2?/2  +  12y=0 

y* 5y— 4  ^2-J-6l/= 4 

2/2+2«y=2a+l 
y-fa=±(a+l) 


44  KEY  TO  ALGEBRA. 

(6.)  Put  x=vy ;  then  the  equations  become 

vy2-\-2y2=60 
56 

y2- 


y2=- 


v2-\-v 
60 

v+2 

15          14 
Hence = 

w-f-2     v2+v 

\5v2  +  15v=Uv-\-28 

15v2+v=28 

4-152t;2+^4-l  =28-60-)-l  =  168l 

2-]5v+l  =  ±41 v=f  or— J. 

(7.)  Put  x=vy  ;  then  the  equations  become 

6i;2i/2+27/2--=5^24-12 (1) 

2ut/2-f3i;2?/2=3i/2—  3 (2) 

12 
From  (1),    .  .  .  y2  = — ■ — 

—3* 

From  (2),    .  .  .  y 


2 

3^2  4-2^—3 


4  1 

Hence  .  .  . j =0 

6u2—  5v+2     3?;2+2v— 3 

12v2+8v—  12+6t;2—  5u+2=0 

18v2+3i>— 10=0 

9v2-t-|v=5 
Complete  the  square,  (Art.  99.)  §v2-\-\v-\-t2=t2-\-5 

3vt=lv     or, t=%     *2=TV 

9v24-Iv+T,a=T,6+5=fi 

3i;-f-*=±} v=f0r— f, 

(§.)  Put  x=vy  ;  then  the  equations  become 

3v2y2+vy2=68 (1) 

4?/2-f-3u2/2  =  l60 (2) 

* 


QUADRATIC  EQUATIONS.  45 

68  160 


3v2+v  "       4+3i> 

40  17 


4+3v     'Sv2-\-v 

120v2+40v=51v+68 

120v2— llu=68 
4-(120)2?;2— .tf+121=68-480+121=32761 
2-120v— ll=rbl81 u=|  or— jj. 

(Art.  111.) 

(2.)  From  the  second  equation  we  get 

x*y*=6  or,         2aA/*  =  12 

This  added  to  the  first  equation  gives 

Put  x*-\-y*=s;  then     s2+2s=35  .  .  .  s=5  or  —7. 

1  1 

Now  if  we  put  a?3=P,  and  y*  =  Q,  we  shall  have 

P+<?=5     and     PQ=6 

From  which  P  and  Q  are  readily  found,  and  from  them 

x  and  v. 

A  A 

(3.)  Put  x*=P,  and  y3  =  Q.     Then  the  equations  be- 
come P+£=8 (lx; 

/"+e"+i*G"=259 (2, 

Square  (1)  and  we  have  P2+2P£+Q2=64  ...  (3) 
Subtract  (3)  from  (2),  and  we  have  P2#2— 2P#=195 
Hence PQ=15  or  —13. 

Now  as  we  have  P+#=8,  and  PQ=-15,  we  have  P= 

1 
5  or  3,  and  #=3  or  5.     That  is,  x*=5  or  3,  &c. 

2.  1 

(4.)  Put  a?3=P,  and  y5  =  #;    then  the  equations  be- 

tome    JF+#2+i>+#=26,     and    P#=8 

2jPQ  =16 

(P+£)2+(P+£)=42     Hence  .  .P+£=6. 


46  KEY  TO  ALGEBRA. 

i 

£C*  33              3             11 

(5.)  Put  — =u  ;  then    u2-\-4u— — .  .  u=-  or . 

yk  4              2             2 

The  remaining  operation  is  obvious. 


(6.)  Given  #2— 8#2y-64,  and  y—2x*y*=4,  to  find  a; 
and  y. 

To  both  members  of  the  first  equation  add  16#,  and  to 
the  second  add  x,  to  complete  the  squares ;  then  extract 
square  root,  and  we  have 

y— 4ar*=4(a?+4)*     and     y*— x*  =  (x+4)* 
Four  times  the  last  equation  subtracted  from  the  prece- 
i 
ding  gives  y — 4i/3=0     or y=16. 

(7.)  Multiply  in  the  first  equation  as  indicated,  and  sub 
tract  the  second  equation ;  we  then  have 

x+y+2x*y*=25     or    x*-\-y*=5 
But  from  the  second  equation  we  have 

(•*-+y*)**y*a-80     Hence #V=6- 

3  3  1 

(8.)  Divide  the  first  equation  by  y2,  and  x3=2y*,  or 
y  =kx3     This  put  in  the  second  equation  gives 
Sxt— £a?l=l4 
al—16#*+64=64— 28=36 
CHAPTER  III. 

QUESTIONS    PRODUCING    QUADRATIC   EQUATIONS. 


We  pass  to  the  sixth. 

(6.)  Let  t=  the  time  (hours)  he  traveled,  and  r=  his  rate 
per  hour;  then  rt=36 (1) 


QUADRATIC    EQUATIONS.  47 

But  if  r  becomes  (r-J-1),  t  must  become  (t — 3),  and  then 
(r+l)(/— 3)=36  .  .  .  . (2) 

Or rt-\-t— 3r— 3=36 

rt  =36 

*.=3(r4-l) 

Hence r2-f-r=12,  and r=3. 

(7.)  Let  x=  the  number  of  children, 

And  •  •  •  y=  the  original  share  of  each. 

Then  .  .  .   a?y=46800 (1) 

(a?— -2)(i/+1950)=46800 (2) 

a?*/-f  1950a?— 2y— 2- 1950=46800 
1950(*— 2)=2i/ 

Or 975(a?— 2)*=a?f*/=46809 

By  division,  3* — 2a?=48 #=8. 

(8.)  Let  a?=  the  number  of  pieces. 

675 

Then =  the  cost  of  each  piece. 

x 

675 

48^. =675 

x 
48a?2— 675a*=675 
16a?2— 225a?=225 


^9.j  Let  a?=the  purchase  money. 

104a?  104a? 

Then  =  the  cost,  and  390 =  his  whole  gain. 

100  100  6 

104a?  104a?  a? 

Then     :     390 :  :     100     :     — 

100  100  12 

Product  of  extremes  and  means 

26a?2 

=39000— 104a? 

300 

2a?2 

=3000—8a? 

300 


48  KEY  TO  ALGEBRA. 

Put  a=300  and  divide  by  2  ;  then 

#2 

— =5a — 4a? 
a 

#2-J-4a#=5a2 
x2+4ax-\-4a2=9a* 
#-J-2a=3« #=0=300. 

(lO.)  Put  x-\-y—  the  greater  part, 

And    •  •  .  x — y  =  the  less  part. 

Then  2#=60,  #=30,  and  x2— i/2=704. 

(11.)  Let  x—  the  cost ;   then  39 — x=  the  whole  gain. 
x     :     39— x     :  :     100     :     x        Ans.  a?=10. 

(12.)  Let  {x — 20)=  the  persons  relieved  by  B. 
Then .  •  .  #-f-20  =  the  persons  .......  A. 

1200  1200 

+  5: 


ar-f-20  #—20 

Divide  by  5,  and  put  «=240  ;  then 


a  a 

1=- 


#+29           x— 20 
o#__20a-f  a?— 400 =ax-\-20a 
#2=4O0+4OO=4O(«-|-1O)=4O-25O 
Or #2=400-25 #=20-5=100. 

Hence  80  is  B's  number,  and  120  A's. 

(13.)  Let  #=  the  price  of  a  dozen  sherry, 
And  y=  the  price  of  a  dozen  claret. 

7x-\-l2y=50 (1) 

10 

— =  the  number  of  dozen  of  sherry  for  ^10 
# 

6 

-=the  number  of  dozen  claret  for  £6, 

y 

10           6 
Then    .  .  .  — =3-f- * (2) 

#  y 


Or 


QUADRATIC  EQUATIONS.  4tf 

10  10y 

70y 
By  substitution,    --- {-12y=50 

70t/+36?/2+72?/=150?/+50'6 

36*/2—  8t/=300 

9#2— 2y=75     Hence y=3. 

(14.)  Let  19#=  the  whole  journey. 
Then  x=  B's  days,  also  his  rate  per  day. 
Or    x2=  B's  distance. 
Also,  7#-{-32  =  A's  distance. 

x2-|-7aH-32  =  19# 
tf2__12#=— 32 

Hence #=8  or  4. 

And 19a?=152  or  76. 

If  we  put  x  for  the  whole  journey,  we  shall  obtain  the 
13th  equation,  (Art.  104.) 

(15.)  Let  x=  the  bushels  of  wheat, 

And    .  .  .  #-f-16=  the  bushels  of  barley. 

24_    24        1 

x     a?+16     4 

x2  +  \§x 

24*-f  16-24=24aH 

4 

a?2-f-16*=16-96=16-16-6 
Put  2a=16.     Then  2a-2«-6=24a2 

x*+2ax=24a2 

x+a=±:5a #=4a=32. 

(16.)  A  put  in  4  horses,  and  B  put  in  x  horses. 

18 
Then  — =the  rate  per  head. 
x 

4 


50  KEY  TO  ALGEBRA. 

4-18 

J— 18==  the  price  of  the  pasture. 

x 

4-20 

__— }-20=  the  price  of  the  pasture. 

x~\~2 

4-18      4-20 
Hence  ....  = \-2 

x        a?-f-2 

36       40 

— =-3-:+l *=6 

x     x-\-2 

(17.)  Let  4x=  the  price  per  yard, 
And    .  .  .  9x=  the  number  of  yards. 

36a?2  =324 x=3. 

(18.)  Let  \0x-\-y=  the  number. 

Then Z£=2 (1) 

xy 

And 10a?+y-r-27==10y-r-a? (2) 

From  (1).  .  .  .  10a?=(2a?—  \)y 
From  (2)  •  •  •  •  x-\-3=y 

^     ,.  .  .  l0x 

By  division,    .  . =2x — 1 

ir-f-3 

1 0a7— 2a?2 + 6a?— <r— 3 

2*2— 5a=3 ^3. 

(19.)  Let  (.r— y),  a?,  and  (a*-}-*/— 6)  represent  the  num- 
bers.    Then  3a? — 6=33,  or  a?=13. 
(x— y)a  —x2 — 2xy-\-y* 
x2=x2 
(x+y— 8)2=x2+2xy-\-y2— 12a?—  12i/+36 

3a?2+2?/2— 12r— 12i/+36=441 

By  subtracting  the  value  of  3a?2 — 12a?+36,  we  have 
2y2—l2y=54:     Hence i/=9. 

(20.)  Let  x  and  y  be  the  numbers. 


QUADRATIC    EQUATIONS.  51 

Then x+y^xy (1) 

X+y=X2+y2    .    .' (2) 

Put  x=vy,  and  the  equations  become 

vy+y=vy2 (3) 

vy+y=v2y2-\-y2    •  • . (4) 

v2  +  l 
Divide  (4)  by  (3),  and   =1 

This  gives  v==h±h  J — 3 

But  from  equation  (3)  we  have  i>y=v-H=«=J±$«/ — 3 

(21.)    Let  x-\-y=  the  greater  number, 

And  ....  x — y=  the  less. 

^^7^24 (1) 

2x+2x2+ty2=62 (2) 

Or x+x2+y2=3l 

Add  (1) x2— y2=24 

2x2-\-x  =55  #=5,     y=l. 

(22.)  Let  x-\-y=  the  greater  number, 
And  .  -  .    x — y=  the  less. 

**— 2/2+2x=47      (1) 

2x2+2y2— 2#=62 (2) 

2x2— 2?/2-r-4a?=94 

4^-f2x=  156 

2x*+  x=  78  x=6,    y=l. 

(23.)    Let  (a?+y)=  one  number, 
And  ....  (x — y)=  the  other. 

2z=27 (1) 

x*+3x*y+3xy*+y* 
Xs — 3x*y  -f-  3.ri/2 — ty3 

2x3+6^2=5103 (2) 

Divide  (2)  by  (1),  and  we  have   **-f-3i/2=189 


52  KEY  TO  ALGEBRA. 

32-92 

\-3y2=9-2l 

3*81 

Divide  by  3,  and f-y=3-21 

4 

3-81-j-4t/2=3-84 

4?/2=3-3     or    2y=3    y==|. 

*-h/=v+f=i5,  &c. 

(24.)    Let  #+y=  one  number, 
And  ....  x — y=  the  less. 

2x=  9 

x4-\-  4x*y-\-  6xy-\-4xf-\-y* 
x4— 4#32/+6#y— 4ay-r-y4 

2a?4+12a?y +2^=2417 
By  resolving  this  we  shall  find  y=\. 

(25.)    Let  #+?/=  the  greater,  and  x — y=  the  less. 

Then O^2— 3/2)  (2#2-f  2*/2)=1248 (1) 

Or x4— -y4=624 

Also 4a??/=20 (2) 

Or y=- 


x 


x 
625 

625 


a?4— =624 

«4 


X* 

#8— 624a?4=625  Put  2a=624 

Then x*—2ax4-\-a2=a2-\-2a+l 

x4— «=±(«+l) 

a?4=2«-f-l  or  — I a?=5 

y=i. 
(26.)    Let  a?=  the  days  required  by  one, 
And  ....  #+10=  the  days  required  by  the  other. 


ARITHMETICAL   PROGRESSION.  53 

Then  .  •  .  -=  one  day's  work  of  the  .first. 
x 

=  one  day's  work  of'  the  second. 

#+10  J 

x    #+10     12 


SECTION     V. 

CHAPTER  I. 
ARITHMETICAL    PROGRESSION. 
(Art.   116.) 

(4.)  Z=l+3kZ (A) 

280=16(24-31)^ (B) 

(B)  reduced  gives  (/=£  ;  then  (A)  gives  Z=16£. 

(5.)  Here  a=|,  L=h  n=5. 

*=l+4d (A) 

*=(*+«i (2*) 

From  {A)  we  have  d=£T 

5+2^=5=  the  first  mean,  &c. 

(6.)  Here  a=9,  Z=109,  n=ll. 

J09=9-f-10rf </=10,  Am. 

(7.)  Z=1+364X2 (A) 

s=730X365X£=(J5),  Ans. 

(8.)  Z=20+(n—  l)3=17+3n (A) 

s=(37-f-3n)irc=438 (#) 


(Art.  117.) 

(2*)  Represent  the  numbers  by  x — y,  x,  and  tf+y. 
3#=18  a?=6 


W  KEY  TO   ALGEBRA. 

x*=x2 
(*4-y)2=*24-2ay-f-*/2 

3x2+2/=158 

(3.)  Let  a?— 3y,  a;—?/,  ar+y,  and  a?-f-3y  represent  the 
numbers ;  then  2?/ =4. 

The  product  of  the  1st  and  4th  is 

a?2-— 9y2 ;  of  the  2d  and  3d  is  (a?—v*). 
x*—y*  v        * ; 

x4 —  9#y 

—  a-y+y 

a?4—10a:y+9y4=  176985 

9^=        144 

x4— 40a?2  =176841 

(4.)  The  same  notation  as  in  last  example. 

2#=8  x=4  #2_y2=i5 

(5.)  Let  n=  the  number  of  days. 

Then Z=l  +  (rc— i)i=n 

*===(l-f»)in=  the  whole  distance. 

Also (n— 6)15      =  the  whole  distance. 

n2-frc=30rc— 180 
n2— 29n=— 180    ....  n=9  or  20 

6         6 

3       14 
(6.)     The  first  day  he  must  pay  1+t;   %  representing 
the  interest  of  one  dollar  for  one  day. 

First  day 1  -j-     { 

2d  day 1 -f-   2i 

3d  day 1+  3i 

Last  day l+60i 

(2+61  i  )30=  the  whole  sum 


GEOMETRICAL  PROGRESSION.  55 

to  be  paid ;  but  as  this  sum  is  to  be  paid  in  60  equal  pay- 
ments, each  payment  must  be 

1-1 —  Ans.  $1  and  f  of  a  cent,  nearly. 

z 

($.)  Let  x — 3^,  x — y,  x-{-y,  and  x-{-3y  represent  the 
numbers;  then  2x2-\-lSy2=50 

2x2-{-  2#2=34 

16*/2=16 y=l. 

GEOMETRICAL  PROGRESSION  AND  HARMONICAL   PROPORTION. 

(Art  124.) 

(1.)  Let  x  represent  the  mean  sought. 

2-6-12 

x= =8 

18 

(2.)  Let  a?=  the  number  sought.     Then,  by  harmonical 

proportion         234     :     x     :  :     90     :     144 — x 

90#=234-144— 234# 

324^=234-144     Hence #=104. 

(3.)  Let  x=  the  number  sought. 

Then    ....  24     :     x     :  :     8     :     4— x 

Or 3     :     x     :  :     1     :     4—  x    .  •  #=3. 

(4.)  Let  x=  the  second. 

Then.  .  .  16     :     2     :  :     16— a?     :     1  .  •  .  .#=8. 


(5.)  Let  x=  the  first  number,  and  y=  ratio. 

Then •  •  x+xy-\-xy2=2l0 (1) 

xy2—x=  90 (2) 

By  subtraction    ....  2x-{-xy=l20     or    x—. 


120 

90 
From  (2)  we  have x= 


56  KEY   TO  ALGEBRA. 

4  3 


— ; — = or,    Ay2 — 3v=10 v=2. 

2+y     y*—l  *        y  y 

(5.)  Let  x,  xy,  xy2,  and  xy*  represent  the  numbers. 

mi  xy3  y2        4 

Then  .  .  .    - — ==-*_=_ 

xy-\-xy2     l-\-y     3 

From  this  equation  we  perceive  at  once  that  y=2  ;  then 

#-r-2#-f-4.r-|-8;r=15#=30 x=2. 

(6.)  Let  x,  xy,  xy2  and  xy*  represent  the  numbers. 

x+xy2=l48 (1) 

xy-\-xy2=888 (2) 

Or x(l+y2)=4-37 (3) 

xy(i+y2)=4'222 (4) 

Divide  (4)  by  (3),  and  ....      y=6. 

(T.)  Let  x,  Jxy,  and  y  represent  the  numbers  ;  then 

x-\-,Jxy+y=14 (1) 

And x2+xy+y2=84 (2) 

Put  x-{-y=s,  and  Jxy—p ; 

Then  •  •  a?-\-xy-{-y*=82 — p2,  and  equations  (1)  and  (2) 

become s-\-p=\4 (3) 

s2—p2=84 (4) 

Divide  (4)  by  (3),  and  we  have     s — p=6 (5) 

Add  (3)  to  (5),  and  divide  by  2,  and 

s=10.     Hence  p=4. 
(8.)  Let  x,  xy,  xy2,  and  xy*  represent  the  numbers  ; 

Then xy3 — xy=24 

xyz-\-x     :     xy2-\-xy     :  :     7     :     3 
Or  .  .  .   y«+l     :       y2+  y     :  :     7     :     3 
Divide  the  first  couplet  by  (y-\-l),  and  we  have 
2/-2/-H      :     y     :  :     7     :     3 
3y2—3y-\-3=7y     or     3y2—l0y=—3 
From  this  equation  we  have  y=3,  the  ratio. 


GEOMETRICAL  PROGRESSION.  57 

(9.)  Let  x,  xy,  xy2,  and  xyz  represent  the  numbers ; 

Then *(i+y+yHy)=y+i 

And a?=Ty  Put  (y+\)=Jl. 

Then TV(^+%2)=^ 

A+Ay*.  =  10./?  rfy2=9A     or i/=3. 

Hence  y'^,  T3o,  &c.  are  the  numbers. 

(10.)    Let  x, ,  and  y  represent  the  numbers;  then 

x+y 

x+—±+y=2G 1 

x+y 

And xy=72 

Put  x+y—s;  then  equation  (1)  becomes 

144 

s-f- =26     or     52—26s=— 144.  ..  .3=18. 

5 

(11.)  Let  x,  xy,  and  xy2  represent  the  numbers ; 

Then «y=216 (1) 

x2+xay4=328 (2) 

From  (I) xy  =  G,     or      x =— 

J/ 
328 

From(2) *~7+y 

36       328  9         82 

or 


y'      l+y*  y2      l+y* 

9^4 — 82y2^ — 9     Hence y=3. 

(12.)  Let  x,  Jxy,  and  y  represent  the  numbers  ;  then 

x+J^y+y=i3 (i) 

(*+y)J*y=*o_ (2) 

x+y=\3—Jxy (3) 

*+y— ^ W 

Jxy 

_       30  — 

13 — Jxy=—=    Hence  .  .  Jxy=3 

Jxy 


58  KEY  TO  ALGEBRA. 

2xy 
(13.)  Let  x,  ,  and  y  represent  the  numbers  ;    then 

x+y 

*+</=18 (1) 

— -=576 (2) 

18  V  ; 

xy 

—  =24         xy=72 (3) 

3 

From  (1)  and  (3)  we  find  x  and  y. 

(14.)  Let  x,  xy,  and  xy2  represent  the  numbers ;  then 
{xy2 — xy)  {xy — x)  are  the  1st  differences,  and 
xy2 — 2xy-\-x=  6 
xy2-\-  xy-\~x=42 

Difference Sxy       =36  ....  .  xy=\2. 

2ry 

(15.)  Let  x,  ,  and  y  represent  the  numbers.     If  y 

x+y 

(2xy  \  /  2xy        \ 
y )  ( x  ) 

4xy 

are  the  1st  differences,  and  y \-x=2  the  2d  diff. 

x+y 
xy=72  Put  (x-{-y)=s; 

4-72 

Then s =2 

s 

s2—2s+ 1=289 

5—l  =  17.  .  .  .s=x+y=18. 


(lY.)  Let  x2,  xy,  and  y2  represent  the  numbers  ;  then 
x2-\-xy-\-y2=3\,     and     x2+y2=26 

(18.)  Let  x,  xy,  xy2,  xy3,  xy4,  and  xy6  represent  the 
numbers.     Then,  by  the  conditions,  we  have 

x-\-xy-\-xy2-\-xy*-\-xy4-\- xy5=lS9=a    •  •  (1) 
And      xy-t-xy*=54=:b (2) 


PROPORTION.  59 

But  equation  (1)  may  be  put  into  this  fofm 

a 
Or x-\-xy3= 

Multiply  this  last  equation  by  ?/,and  its  first  member  will 
be  the  same  as  the  first  member  of  equation  (2) ;  therefore 

ay 

i+y+3/2 

ratio. 


•■=&  ;  a  quadratic  from  which  we  obtain  y,  the 


(19.)  Take  the  same  notation  as  for  (18) ;  then  we  have 

{x+xy)+(x+xy)yA  =  \S$— 36=153=«.  .  (1) 
And \x+xy)y*=ZQ=b (2) 

Divide  (1)  by  (2),  and  we  have 

1+V4      153     51      tt 

£_= =_     Hence y=2. 

y*  36       12  * 

CHAPTER  III. 
PROPORTION. 


(5.)  Let  x  and  y  represent  the  numbers  ;  then 
x — y     :     x-\-y     :  :       2     :       9 
x-\-y     :       xy       :  :     18     :     77 

From  the  first,  2x     :     2y     :  :     11     :       7or,a?=yy. 


ISy  U^ 

7  7 


18     :     77 


y     :     Uy*     :  :       1     :     77        y=7. 
(6.)  Let  x  and  y  represent  the  numbers. 

x+4     :     y+4     :  :     3     :     4    ....  (1) 
#_4     :     2/_4     :  :      I     :     4     ....  (2) 

From  (2)  we  have  4a? — 16=y — 4,  or,  y=4x — 12.  This 
value  of  y  put  in  (1)  gives 


60  KEY  TO   ALGEBRA. 

x-\-4     :     4a;— 8     :  :     3     :     4 
x-\-4     :       x — 2     :  :     3     :     1 

x+4=3x—  6 x=5. 

(7.)  Let  x  and  y  represent  the  numbers ; 

Then x-hy=lG 

And xy     :     x*-\-y2    :  :     15     :     34 

Double  the  first  and  third  terms,  then  add  and  subtract 
(Theorem  4),  and  2xy     :     x2-\-y*     :  :     30     :     34 
x*i-2xy-\-y2    :     x2— 2xy+y2     :  :     64     :       4 
x-{-y     :     x — y     :  :     8     :     2 

16     :     x — y     :  :     4     :     1    or,  a: — y=4. 

(8.)   Let  #=  the  gallons  of  rum, 

And  .  .  .  y=  the  gallons  of  brandy. 

x — y     :     y     :  :     100     :     x 
x — y     :     x     :  :         4     :     y 

Product     (x — y)2    :    xy    :  :     400     :    xy 
Dividing  the  second  and  fourth  by  xy,  and 

(x—y)2     :     1     :  :     400     :     1 

x — y       :     1     :  :       20     :     1  or,  a;— y=20. 
(9.)  Let  x-\-y=  the  greater  number, 
And  .  .  •  x — y=  the  less. 

Then    x9—y2=320 (1) 

{x+yf^+Sxty+Sxy'+y* 
(x — y  )3 = xs — 3x*y + 3  xy* — y3 

6#?3/+2?/3=diff.  of  the  cubes. 
2y=  difference.     Cube  of  (2?/)=  8I/3 
§x2y+2y*     :     8ys     :  :     61     :     1 
3X2  -f-  y2     :     4y2     :  :     61     :     1 
3x2+y2=244y2  3a*=243y2 

^=  Sly2 
This  value  of  x2  put  in  equation  (1),  gives 

80t/2=32Q    or y=2. 


PROPORTION.  61 

(10.)  Let  a?  and  y  represent  the  numbers ; 

Then a?-f-t/=60 

And  .  .  .  .  xy     :     x2-\-y2    :    :    2     :     5 
Double  the  first  and  third  terms,  aud 

2xy     :     x2-\-y2    :    :    4    :    5 
Add  and  subtract  and 

x3-{-2xy-\-y2    :    x2 — 2xy-\-y2    :  :    9    :    1 

60     :     x—y     :    :     3     :     1  x—y=20 

But a?+y=60 

(11.)  Let  3a?  and  2a?  represent  the  numbers; 

3a?-}-6     :     2a?— 6     :    :     3     :     1       Hence,  x=8 

(12.)  Let  16a?  and  9a?  represent  the  numbers ; 

Then  .  .  16a?     :     24     :    :     24     :     9a?  Hence,  a?=2 

(13.)  Let  a?  and  y  represent  the  numbers  ; 
Then  x-\-y     :    x — y    :  :     4     :     1         Hence,  5y=3x 

This  reduced  equation  will  be  true  if  we  take  y=3  and 
a?=5,  or,  if  we  take  any  multiples  of  5  and  3,  as  10  and  6, 
15  and  9,  20  and  12,  &c.  Hence  the  answer  in  the  book 
is  correct,  but  many  other  answers  are  equally  correct,  and 
the  problem  is  therefore  indeterminate. 

Again  let  a?  and  y  represent  the  numbers  i 

And     a?2 -ft/3     :       x       :    :     102  :    5 

Or,  a?2 +29^ x2     :       a?       :    :    102  :    5  Hence  a?=  15 

(14.)  Let  a?  and  y  represent  the  two  numbers 
Then,  x-\-y=20,    and    a?     :    y     :    :     9     :     1 


62  KEY  TO  ALGEBRA. 

APPLICATION  OF  THE  BINOMIAL  THEOREAL 

(3.)  To  expand  (a — £>)~',we  take  formula  (3).     Then 
a=a,  x= — 6,  and  m— — 1. 

Hence, 
am+mam-'b-\-mPj  am~2b2,  &,c.=a~x  +a--6+a"362,  &c. 
"  1      b      b2 

Or -+-7+-T'  &c- 

a     az     cr 

b2 
(4.)  Take  formula  (1),  and  put  x= — .     Then 

a2 

K         J  \  2      2  22  3  / 

/        b2        b*  368  \ 

a(  H H &c.  ) 

\       2a2     2-4a4     2-46a6        / 

(5.)     Expand = -=-(  H ) 

(«•+*•)*  c(n-gy   A    c2) 

The  numerical  coefficients  of  this  series  must  be  the 

same  as  the  last,  because  m  is  numerically  the  same  ;  the 

sign  of  the  second  and  every  alternate  term  will  be  minus. 

d/        x2       3x4        3-5x6        3-5-7.T8  \ 

-(  1 H -f- Ac.  ) 

c\        2c2      2-4c4     2-4-6c6     2-4-6'8c8        / 

,     X3/       «2\?       |/       #2\f      a2  /       a*\f 

(«•)  (*)'0— )  =<<'—)  =70(1^) 

3  a;        x2 
Here m= 


4  a         a2 
x         m — 1    x2 

Formula  (2)     1+m — \-  m- &c.  = 

w  a  %       a2 

/       3^      3    a*4        3  5*6  \ 

(  1 <fcc.  )      This  multiplied   by 

V       4  a2     4-8  a4     4-8-  12c^        /  *  , 

a2  /l/        3^      3x4       5x6        \ 

the  factor will  give  \/  -l  a2 &c.  1 

j  a  a\         22     25a2      27c4        / 


COMPOUND  INTEREST.  63 

(7.)  (a+y)m+=a"+mar-'y-\-m'?i±am-2y2J  &c,  and 
il  m== — 4,  the  series  will  be 

a--^— 4a-^/-r-10a-y—  20a-y'-f-  &c. 

1      4?/     l(ty3     20  w3 
Or -H ■ &c. 


a3  1         /       63V' 

(8.) = =(  H }      The  cube  root  of  this 

V    '  a3+63  IP     \       a3/ 

a3 

is  .  .  .(l-h— )       =1 1 &c. 

V       a  /  3a3     3*6a6       3-6-909 

63      266       1469 
Or =1 h &c. 

3a3     9a6       81a9 

COMPOUND   INTEREST. 


(4.)  The  general  equation  is  pJin=a,  and  for  this  exam- 
ple jo=5,  c#=(1.05)  a=9,  and  n  is  unknown. 
Hence 5(l-05)n=9,    or    (i*05)n=l-8 

log.    1.8        0.25527 

Or  •  .  .  .  n= = =12.04  years,  nearly. 

log.  (1-05)     0.02119  J  ' 

(5.)         1000(l+r)6=1800     or     (l+r)6=1.8 

By  logarithms  ...  6  log.(l+r)  =  log.  (1.8) 

0*25527 

Hence  .  .  .  -log.  (l-fr)  = =0.04254 

6 

By  the  tables  we  find  l-{-r= 1*103,  or,  r=.10T\,  nearly. 

(6.)  In  this  example  the  general  formula  is 

jo(1.04)4=350.9575.       By  logarithms  we  have 
log.  p+4  log.  (1.04)=  log.  350.9575 

Or log.  /H-0.06812=2. 54525 

By  reduction log.  jo=2. 47713 

Hence jo=300,  Ana. 


64  KEY  TO  ALGEBRA. 

The  general  equation  applied  to  this  problem  is 

3600(1.05)"=5000(1.04)1 2 

By  reduction  .  .  .  .  (1.05)7,=     |°(1.04)12 

n.  log.  (1.05)  =  log.  fl+12  log.  (1.04) 

w(0.02119)=0.14267+0.19236 

0.33503 

n= =16  nearly. 

0.02119  J 

ANNUITIES. 

(6.)  That  is,  the  rent  is  an  annuity  to  continue  forever ; 

what  sum  of  money  will  purchase  it  ? 

mi  p     3000 

The  general  equation  is  P=-= =100000. 

r      .03 

(7.)  The  general  equation  applied  to  this  problem  is 

350[(1.04)8— 1] 
JSt'= — -=8750[  (1.04)8— 1] 

Log.  J2'=  log.  8750+  log.  ( (1.04)8—  1) 
But (1.04)8— 1=.3685 

Log.  .#'=3.94201— 1. +0.56644=3.50845 

A' 
But  the  answer  to  the  question  is  P= —     (Art.  155.) 

That  is  by  log.  (log.  P.) =3.50845—0. 13624=3.37221 
Or P=2356.46  Jlns. 

(8.)  If  no  interest  were  required,  the  sum  to  be  paid  at 
each  annual  payment  would  be  t3°°  dollars ;  but  this  must 
be  paid  and  compound  interest  on  the  same  for  1,  2,  3,  &c. 
years,  up  to  7.     Call  —. 7—  =/?. 

Let  ^=(l+r)  =  1.04.     Then  by  (Art.  153), 

The  sum  to  be  paid  the  first  year  must  be pA 

' second  year pJP 

• third  year pJP 

• last  year p^F 


EQUATIONS.  65 

This  is  a  geometrical  series,  and  its  sum  (Art.  120)  is 

A — 1  r  r  ' 

1200     1.04/                  N     300.104 
Or,  5= X (  (1.04)7— 1  )= X.3158 

7         .04  V        '        J  7 

But,  if  this  sum  is  to  be  paid  by  seven  equal  payments, 

each  payment  must  be 

300X104  312 

X.3158= X31.58=$201-l-  dns. 

49  49 

p     250 
(9.)   Here  P=-= =$35712  Jlns. 

v    '  r     .07  7 


SOLUTIONS  OF  EQUATIONS  OF  THE  HIGHER  DEGREES. 
NEWTON'S  METHOD  OF    APPROXIMATION. 

(1.)  Given  x3-{-2x2 — 23# ^70,  to  find  one  value  of  x. 

By  trial  we  find  that  one  value  of  x  is  between  5  and  6, 
nearer  5  than  6  ;  therefore,  let  a=5  and  y—  the  remaining 
part  of  the  root.     Then  x=a-\-y. 

Expand,  neglecting  all  the  terms  containing  the  powers 
of  y  after  the  first,  and  we  shall  have 

xz=  a3-f3a22/-f  &c. 
2x2  =2a2-\-4ay+  &c. 
— 23x=— 23«— 23y 

By  addition, 

ar4-2;c2— 23#=«H-2«2—  23a-f-(3a2+40— 23)y=70 

In  this  last  equation  we  observe  that  a  has  the  same 
powers  and  coefficients  as  x,  and  the  coefficients  to  y  may 
be  found  by  the  following 

Rule.  Multiply  each  coefficient  of  x  by  its  exponent, 
diminish  each  exponent  by  unity,  and  change  x  to  a. 


66  KEY  TO   ALGEBRA. 

70-|-23a— 2a2— a3 

Now    y  — — Giving  a  its  value,  5,  we 

3  Sa2-\-4a— 23  6 

have    y=s||a=.l4-       Now  make  a=5.1,  and   substitute 

again  in  the  preceding  formula,  we  have  a  new  value  of  y. 

2.629 

Thus  y= =.03   Now  make  a^5. 13,  and  substitute 

u     75.43 

again,  and  our  new  value  of  y  will  be  .004578-}-     Hence 

a+y  or  #=5.134578-{- 

(2.)  Given  x4— 3a?*-f  75^=10000,  to  find  one  value  of  x. 

By  trial  we  find  x  must  be  near  10.     Hence  put  a=10 

and  x=a-\-y.     Then  by  the  preceding  rule 

10000— 75a-{-3a2—  a4      —450 

v= =  =—.11 

9  4ft3— 6a-r-75  4015 

Now  make  «=10 — .11  =  9.89.  If  we  have  the  patience 
to  substitute  this  value  for  a  in  the  equation,  we  shall  have 
a  new  value  to  */,  true  to  6  or  7  places  of  decimals,  and  of 
course  a  value  to  x  to  the  same  degree  of  exactness. 

(3.)  Given  3x4— 35x3—  11a.-2—  14#-f-30=0  to  find  one 
value  of  x. 

By  trial  we  find  that  x  must  be  near  12.  Let  a=12, 
and  x=a-\-y.     Then  by  the  rule 

— 30+14H-lla2+35a3— 3a4     —6 


y- 


-00112 


12a3— 105a2— 22a— 14  5338 

Hence z=12— .00112=11.99888. 

(4.)  Given  5x? — 3x2 — 2.r=1560,  to  find  x. 
We  find  by  trial  that  one  value  of  x  is  more  than  7.    Put 
x—a-\-y,  and  a=7.     Then  by  the  rule 

1560+2a4-3a2— 5a3       6 

y= ! = =.00867+ 

*  15a2— 6a— 2  689 

Hence #=7.00867+ 


EQUATIONS.  67 

CHAPTER  in. 
YOUNG'S   METHOD  OF   RESOLVING   THE   HIGHER  EQUATIONS. 

This  is  really  Horner's  method,  but  extracted  from  Young. 


(6.)  Given   ^+7^=1194,  to  find  the  values  of  x. 
We  perceive  by  inspection  that  one  value  of  x  must  be 
more  than  31 ;  therefore  put  r=31. 


«-f-r= .  .  .  . 

•  38 

1194  (31.2311-f  &c. 

r+s 

31.2 

1178 

a-{-2r-\-a 

69.2 

1600 

s-\-t 

23 
69.43 

1384 

21600 

31 
69.461 

20829 

77100 

&c. 

69461 

&c. 
As  the  sum  of  the  two  roots  is  equal  to  — 7,  the  other 
root  must  be  — 38.23 11 -f- 

(7.)  Not  necessary  to  have  place  in  a  key. 

(8.)  Given  x2— 21^=214591760730,  to  find  one  value 
of  x. 

In  this  example  the  pupil  might  be  at  a  loss  as  to  the 
most  expeditious  manner  of  finding  r  by  trial. 

Conceive  — 21a?  not  to  exist ;  then  the  value  of  a?  will  be 
the  square  root  of  the  absolute  term ;  but  this  term  has  six 
periods  of  two  figures  each,  and  the  superior  period  is  21. 
The  greatest  square  in  this  is  16,  root  4  ;  hence  r  must  be 
at  least  400000,  six  places  ;  try  this  number. 


68 


KEY  TO  ALGEBRA. 


— a-\-r .... 
r-f-a  •  •  • 

=399979 
.  460000 

214591760730 
1599916 

(  400000=r 
60000=* 

~a+2r-|-s.  .  . 

•  859979 
.  .  63000 

5460016 
5159874 

3000=* 
200 =u 

a+2r+2s+t 

.  922979 
3200 

3001420 
2768937 

50=v 
l=tt> 

926179 
250 

2324837 
1852358 

« 

926429 
51 

4724793 
4632145 

#=463251. 

926480  926480 

926480 
As  the  algebraic  sum  of  the  two  roots  must  make  21, 
(Art.   156  in  the  work,)  therefore  the  other  root  must  be 
—463230. 

(9.)  Given  7x2 — 3#— 375,  to  find  one  value  of  x. 
Or  #2—  ?#=3I5.     Putx=\y.     (Art.  166.) 


Then 


y2     Sy     375 


Or  I/2— 3y=375X  7=2625. 


49     49       7 

In  this  equation  we  perceive  that  y  must  be  more  than 

the  square  root  of  2625,  that  is,  more  than  50.     Hence  put 

r=50. 

rs  t 


— a-\-  r .  .  .  . 

.47 

2625  (  52 

r-f-s 

52 

235 

— a+2r-fs.  . 

.  99 

275 

s+t 

2.7 

198 

1017 

7700  u 

75 

7119  H 

10245 

58100 
51225 

6875 

52.756-f-    „   , 
Hence  x= =7.+ 


EQUATIONS.  69 

(10.)  Given  2x* — lla?= — 7|,  to  find  one  value  of  x. 
Orx2-  'Jar==— y     Put  a?=£y. 
Then  .  .  .  ly2—  y y=— y     or.  .  .#2—  lli/= — 15. 
In  this  equation  one  value  of  y  is  between  9  and  10 ; 
therefore  put  r—9. 

— a+  r —2  —15  (  9.405124838+ 

r-\-s  9.4         —18 


— a-f-2r-f-s  7.4  300 

*+*  4  296 


7.805  40000 

51  39025 


78101  97500 

12  78101 


781022  1939900 

24  1562044 


7810244  377856 

The  division  is  not  carried  out  for  the  last  three  figures. 

Hence  one  value  of  y  is  9.405124838-J-,  and  as  the  two 

values  must  make  11,  (Art.  166),  the  other  value  must  be 

1.594875161.     But  x=iy,  therefore 

#=.797437580+  or,  #=4.7025624194-  Am. 
(11.)  Given  |#2+|a?=T7T,  to  find  one  value  of  x. 
Or  #2-f-ia:=!f  or,     ar2+.8#=0.848484848+ 

Here  it  is  obvious  that  x  cannot  be  1  ;  by  trial  we  find 

it  must  be  near  .6 ;  therefore  r=6. 

a+r  1.4      0.8484848484  &c.  ( 0.604233+ 

r+s   60       84 


2004        8484 
42       8016 


20082        46884 
23       40164 


200843        672084 


70  KEY  TO  ALGEBRA. 

We  omit  several  examples  as  they  present  no  difficulty. 

CUBIC   EQUATIONS. 


(3.)  Given  x*-{-2xa— 23#=70,  to  find  one  value  of  x. 

By  trial  we  find  x  must  be  a  little  over  5;    therefore 
r=5,  .#=2,  ,#=—23,  N=70. 

B —23 

r(r+A) 35") 

1st  Divisor 12  >  70  (  5.134 

r2 .  .  25J  60 

B' 72  10000 

*(a+3r+^) 17n  7371 

2d  Divisor 7371  ^  2629000 

s2 1J  2276697 

B" 7543         352303000 

*(3Q+t)t 4599       305649104 

3d  Divisor 758899        46653896 

f 9 

B" 763507 

61576 

4th  Divisor 76412276 

16 

Common  division  will  give  three  or  four  more  figures  to 
perfect  accuracy. 

(4.)  Given  xs — 17^+42^=185,  to  find  one  value  of  x. 
Here  .#=—17,  5=42,  JV=185,  and  we  find  by  trial 
that  x  must  be  between  15  and  16  ;  therefore  r=15. 

*  Q  represents  the  root  as  far  as  previously  determined. 


CUBIC  EQUATIONS.  71 

B 42 

r(r+Ji) —30^ 

I  r  st 

1st  Divisor   ...   12  >  185  (  15.02 

r2  .  .  . 225J       180 

B' 207         5000000 

*(s-f3r-M)  ....   0")        4154008 

2d  Divisor 207  f       2077  )  845992  (  407 

s2 OJ         8298 

B" 207  16192 

t(3Q-{-t) 7004        14539 

3d  Divisor 2077004        1653.0 

Hence x=15.02407-|- 

(5.)  Given  x3-{-xi-^500i  to  find  one  value  of  x. 

Here  Aml%  B=0,  r=7. 
B 0 

r(r-\-Jl) 56 

1st  Divisor    ....  56  500  (  7.61 

r2 49  392 

B' 161  108 

(3r+*-M)s .  ■  •  •  1356  104736 

2d  Divisor  ....   17456  ■  3264 

*a 36  1887181 

18848  )1376819 

2381 
Continue  by  common  division. 

3d  Divisor  ....  1887181 
1 

1889563 

(6.)  Given  #3+10x2-j-5tf=2600,  to  find  one  value  of  x. 


72  KEY  TO  ALGEBRA. 

Here  ^=10,  B=5,  r=ll. 

B 5  2600  (  11.006 

r(r+^).  ...    23  n  2596 

1st  Divisor .  .  ■  236  V  4 

r2 121J  3529188216 

B'     588  470811784 

(3/?+tt)w.  •  •  •        198036 

.  ,   t. .  .  ~nntnnn^n  Continue  by  common  division. 

4th  Divisor  .  .  .  588198036  J 

The  four  miscellaneous  examples  on  page  262,  the  last 

page  of  the  school  edition. 

(1.)  Let  x-\-y  represent  the  greatest  extreme,  and  x — y 

x2 — y2 
the  least  extreme ;  then  — — =  the  harmonical  mean. 
x 
By  the  conditions  of  the  problem  we  have 

,  x2 — v2  (x2 — V2) 

2x+ -=26.  •  .  .(1)     and     * —=576    .  .  .  (2) 

x  x 

By  reducing  (1),  and  taking  the  square  root  of  equation 

(2),  we  have  2x2+x2— y2=2Qx (3) 

x2—y2=24)Jx~ (4) 

By  subtraction  .  .  .  2x2=     2Qx — 24  Jx 

If  we  put  jjx=v,  and  reduce  the  resulting  equation,  we 
shall  have »*—  13t>=— 12 

By  (Art.  163)  we  find  v=3.    -Hence  a?=9,  &c. 

(2.)  Let  x-\-y=  the  greater  number,  and  x — y=  the  less. 
Then  by  the  problem  we  have    x=5 

And 8^1/4- 8.n/3=1040 

Or y*+25y=26     Hence y=l 

(3.)  Let  x,  y,  and  z  represent  the  numbers  ;  then  by 

the  problem x-\-y-\-z=2S (1) 

x+yz=5l (2) 

y+xz=S7 (3) 


EQUATIONS.  73 

By  adding  (2)  and  (3)  we  have  (x+y)-\-(x+y)z=l38 
But  by  equation  (1),  x-\-y =28 — z  ;  therefore 

28— s+28z— z2=138     or      z=5 

(4.)  Let  x,  y,  and  z  represent  the  numbers  ;  then 

^+2/2+zs=  195 (1) 

x9+y9+z3=l799 (2) 

xyz=  385 (3) 

The  form  of  these  equations  will  not  be  changed  by  ta- 
king x  for  z  or  x  for  y.  A  full  and  formal  solution  of  these 
equations  would  become  tedious,  and  to  avoid  this,  we  may 
venture  a  solution  by  inspection.  As  195  and  385  both 
have  5  in  the  unit's  place,  it  is  more  than  probable  that  the 
value  of  one  of  the  symbols  is  5.  Therefore  assume  z=5. 
This  gives  ic2-f-i/2=170,  and  #*/=77,  from  which  we  find 
a?=7  or  11,  and  y=ll  or  7,  and  as  5,  7,  and  11  will  verify 
equation  (2),  this  is  a  true  solution. 
The  following  examples  are  in  the  University  Edition  only. 

(6.)  Find  one  value  of  x  from  bx*— 6x2-\-3x=— 85. 

As  the  result  is  negative,  we  will  change  the  second  and 
every  alternate  sign  of  the  equation,  (Art.  178),  and  find  a 
value  of  x  from  the  equation      5rJ-f-6x2-r-3a;=85 

Use  the  formula  of  (Art.  194.)  c=5,  J3=6y  J5=3,  and 
by  trial  we  find  r=2.  r  s 

B 3  85  (2.1 

(cr-f-^)r ....    32V  70 

1st  divisor    .  .    35  f  15 

cr* 20J  9.065 

87  5  935 

*  '    *  *  Continuing  this  we  shall  find  the 

2d  Divisor .  •  •    90.65  value  of  x  to  be  2.16399-f ,  and  its 

cs _^ sign  changed  will  be  the  value  of  x 

94.35   in  the  original  equation. 


74  KEY  TO  ALGEBRA. 

(7.)  Find  x  from  the  equation  \2x*-{-x2 — 5#=330 

Here  c=12,  J2=l,  J5=— 5.      r=3. 

£ —5 

(cr+A)r lin 

I  r  »< 

1st  Divisor  ....  106  ^  330  (  3.036 

cr* 108J  318 

B' 325  "~12 

{3cE-\-ct)t.  .  .  •       11208  9783624 

3d  Divisor.  .  .  .   3261208  2216376 

ct2 108 

Continue  thus. 

3272524 

In  the  same  manner  perform  (8)  and  (9.) 

(  Art.  195.  )  Page  323. 

(3.)  Extract  the  cube  root  of  1-352-605-460-594-688. 

For  the  sake  of  brevity,  take  r=ll,  in  place  of  1. 
1st  Divisor.  •  -121  r  st 

JS'^r2.  •    363  1-352-605-460-594-688  (110592 

(3B-\-t)t.  •       16525       1  331 

2d  Divisor    3646525  21  605  460 

25  18  232  625 

3663075  3  372  835  594 

(3R+u)u  •  •      298431         3  299  453  379 

3d  Divisor    366605931  73  382  215  688 

81  73  382  215  688 

366904443 
(3i?-f-v)u  •  ■  663544 

36691108844 

(4.)  By  a  table  of  cubes  which  run  to  8000,  we  perceive 
at  once  that  r  in  this  example  is  17. 


EQUATIONS 

1st  Divisor 289 

Sr2=B' 867 

(3#-fs)s 2575 

2d  Divisor 89275 

25 


(3B+t)t 


9J875 
10504 


rs  t 
5382674  (  175.2 
4913 

469674 
446375 


23299000 
18396008 


75 


3d  Divisor 9198004  4902992 

4      Complete   another   divisor, 
9208512  t*ien  contmue  as  m  simple  di- 


It  is  not  important  to  show  a  solution  to  the  remaining 
examples  under  this  article. 

Page  324. 


(»•) 

Change 

all  the  signs, 

then 

a*+a*+3?— £=500 

r=4 

r\     i 

1 

—1 

500  (4 

4 

20 

84 

332 

5 

2l 

83 

1~68 

c 

4 

36 

228 

9 

57 

311 

go 

4 

52 

b 

Ti 

To9 

• 

4 

_    8 

1     17 


109 


311 


168  (.4  <fcc. 


(2.)  Resolved  in  the  same  manner  as  (1.) 

(3.)  The  signs  of  all  the  terms  must  be  changed,  then 


76 

'1       —9 

.1 

KEY   TO 

—11 
—.89 

ALGEBRA. 

—20 
—1.189 

— 4(.l 
—2.1189 

03 

cJ 

S-l 

-•J 

| 

fa 

—8.9 

1 

—11.89 

—  88 

—21.189 
—  1277 

—1.8811 

—8.8 

1 

—12.77 

—  87 

—22.466 

—8.7 
1 

—13.64 

1       — 

-8.6      - 

-13.64 

—22.466  — 1 

.8811  (.07  &c. 

(4.)  x5= 

5000. 

Here  all  the  coefficients 

are  zero  ex- 

cept  the  first,  and  r- 

=  5. 

r 

1           0 

0 

0 

0 

=5000  (  5 

5 

25 

125 

625 

3125 

5 

25 

125 

625 

1875 

5 

50 

375 

2500 

10 

75 

500 

3125 

5 

75 

750 

15 

150 

1250 

5 

100 

20 

250 

5 

25 

s 

1         25 

250 

1250 

3125 

=  1875  (.4&c 

(5.)       X*: 

64.Z2 

= or 

z4+2r4-l 

ars+2»8+s= 

=64 

r     1 

1             0 

2 

0 

1 

=64(2 

2 

4 

12 

24 

50 

2 

6 

75 

25 

14 

2 

8 

28 

80 

4 

14 

40 

105 

Continue  as  in  last  example. 


EQUATIONS.  77 

In  the  editions  of  Algebra  which  we're  published  prior 
to  1857,  some  few  important  problems  were  passed  over 
in  the  Key  without  a  solution. 

They  were  purposely  omitted,  as  an  experiment,  to  dis- 
cover whether  there  would  be  any  special  call  upon  the 
author  for  solutions;  and  the  experiment  is  now  over.  The 
call  for  their  solution  has  been  very  frequent  and  earnest, 
from  all  parts  of  the  country:  and  this  demonstrates  the 
importance  of  a  Key. 

We  now  insert  them,  and  some  others  which  have  been 
added  to  the  enlarged  edition,  first  published  in  1858. 

A  more  general  Key  to  mathematical  science  is  pub- 
lished in  our  Operations.  In  that  work,  the  advanced 
student  and  the  amateur  mathematician  will  find  many 
curious  and  useful  problems,  not  merely  in  Algebra,  but 
in   Geometry  and  the  higher  sciences. 

(Art.  107.)  Example  18. 

a 

Given  ....    x — l=24-__ 
Jx 

Place  .   Jx=y.         Then  .  y2— 1=2+? 

y(y»— l)=*(y+l) 

Divide  by  (y+1)     then  y(y— 1)=2  or  y2 — y=2. 
Whence, y=2     or     1. 

(Art.  114.)  Example  17. 

Let  x  and  y  represent  the  number.  Then,  by  the  given 
conditions,  we  have: 

x+y=xy         ( 1 ) 

and     xy=x2 — y2  (2) 

Assume  x=vy.     Then  vy-\-y—vy2 ,     or     y=l-| —        (3) 
and     vy2—v2y2 — y2 


78  KEY  TO  ALGEBRA. 

Whence,   .  y=ldn_=_J_ 

But,  .  .   .  •«f-^J  •  L^=H3±^) 

y    l+^5    1-V5   -~i-4        ^1+^;- 

Example  31.     (Page  189.) 
Let  x  and  y  represent  the  number.     Then,  by  the  given 
conditions: 

xy—x2 — ya  (1) 

x*+ya=x*— y3         (2) 
Assume    x=vy.     And  this  value  of  x,  placed  in  (I)  and 
(2),  produce 

vy2=v2y2 — y2  (3) 

and     v2 y2  -\-y2  =v3  y3 — y3         (4) 
Dividing  (3)  and  (4)  each  by  y2 ,  and  we  obtain 
v=v2— 1         (5) 
and     v2+\  =  (v3—l)y         (6) 
Observe  that  (5)  is  a  quadratic,  and  its  solution  gives 

2*>=l±>/5         (7) 
Again:  Add  2  to  each  member  of  (5),  and  we  have 

v-\-2=v2-\-l. 
Also,  multiply  (5)  by  v,  and 

v2=v3 — vy     or     #2~j-z;=fl3. 

But v2=v-\-\\ 

Whence,   .  2v-\-\—v3 ,     and     2v=v3 — 1. 
Now,  Equation  (6)  becomes 
2vy=v-\-2y 
or         4vy=2v-^-4. 
Substituting  the  value  of  4v  and  2v,  as  found  in  (7),  we 
have 

2(\±j5)y=5±j5 

2y=5/y±T=±V5     *******  Ans' 

Lastly,       x^vy=^  (\d-j5)^s/5=\  (J5±5)     Arts. 


EQUATIONS.  79 

Another  solution  may  be   found  in  our   Mathematical 
Operations,  pages  115  and  116. 

(Art.  204.)  Example  1. 

x*—5x3+5x2—l=0. 
Here  the  sum  of  the  coefficients  is  zero,  therefore  one 
root  is  -f-l>  and  the  equation  is  divisible  by  x — 1. 
Dividing  by  (x — 1),  the  quotient  is 

x*-\-x3—4z2-\-x-{-l=:0. 
Here,  again,  the  sum  of  the  coefficients  is  zero,  show- 
ing that  another  root  is  +1;  and  this,  again,  is  divisible 
by  (x — 1).     The  next  quotient  is 

x3+2x2—  2x— 1=0. 
Here,  again,  the  sum  of  the  coefficients  is  zero,  show- 
ing a  third  root  equal  to  -(-1;  and  hence,  we  divide  again 
bj  x — 1.     The  quotient  is 

s*_|-3;z+l=0.         Whence,  x=  —  \  (3±J5). 
N".  B. — We  might  have  taken     x*-\-x3 — 4#2-|-£-|-l=0, 
and  solved  it  by  the  rule  (under  Art.  203). 

2.  x*+5x3~\-2x2±5x-{-l=0. 
By  the  rule  (under  Art.  203)  we  have 

x<  -j_5z3  -f(  2+V  )xa  -j-5^-1-1  =  yz2 ; 
Extracting  square  root,  and 

x2=— 1,         or      x=±J—l; 
Or, x2+5x=— 1  *=H— 5±J2l~ 

3.  x*—  4x3-\-4x— 1=0. 
(a.*— 1)— 4.r(z2  — 1)=0. 

Dividing  by  (x2 — 1),  and  x2-\-\ — 4^=0,  or  x2 — 1=0, 
Or  .   .  #=d=l;         or     x2 — 4.r-)-4=3.  #=2±A/3. 

4.  x*—^x3-{-2x2 — \x  -(-1=0. 

By  the  rule   (Art.  203),  add  \\x2  to  each  member— 
then  *«—  lx3+\\x2—  fz-f^ffz2 

Square  root,  >  x2—  |-ar+l  ==fc|ar, 


80  KEY  TO  ALGEBRA. 

Whence,   .   .  #2-|-l=0,     or     x2 — fa--|-l=0 

x=±z,J — I,     or     x=2      or     £. 

5.  5x*-\-Zx3+9x2  +8a--f-5=0. 

Add  ....  2a--2  =-|-U>a;2 

and     #4+fa:3+2a:2+fav}-l  =     ?L 

Add ISfl  =1^1 

25 25_ 

Square  root,  x2-\-$x-\-l=±xj2\         a  quadratic. 

Place  ....     4~-^21=2fl  4— 721=10a, 

5 

Then  ....  s2+2az+a2=a2— 1 

a?= — a-±ija2 — 1 

Because  a2= — _5l —  /a2 j   is  imaginary,  and, 

therefore,  the  values  of  x  are  imaginary. 

6.  4a*6— 24a?5+67a:4—  73a-3  +57a-2— 24a-+4=0. 
Divide  by  x3;  then 

4a:3— 24ar2+57a;— 73+^— ^4--i-==0  (1) 

XXX 

1  ^7 

Assume  a--|--=y,  Then      57a;+ — =57y, 

a;  a; 

a:a+2+-L=y2  —  24z2— — =48— 24ys 

a;2  a;2 

Also,    a-3+-U-3  (x-\S\^y\     or    a-^*^3— 3y, 
#3  \        a?/  * 

4 

and         4#3  4- — =4y3 — 12y. 
x3 
These  values  placed  in  ( 1 ) 
And  4y3  —  12y+48— 24y2+57y—  73=0 


EQUATIONS.  81 

Or  4y3— 24y2-f-45y— 25=0,  (2) 

Here  the  sum  of  the  coefficients  is  'zero.  Therefore, 
one  value  of  y  is  1. 

Then  &4-!=l.  Whence     x=l±z^~^- 

1  x  2 

Dividing  (2)  by  y — 1,  and  we  obtain 
4y2— 20y+25=0 
Square  root  2y — 5=0,     or  y=f. 

Whence         #-!-_=§,         and    #=2         or      £. 
x 

7.  4j?4+3«3— 8a;2— 3ar+4=0.  (1) 
Here  the  sum  of  the  coefficients  is  zero.     Therefore, 

#=1  for  one  root.  Again,  if  we  change  the  second  and 
every  alternate  sign,  we  shall  have 

and  the  sum  of  the  coefficients  is  still  zero;  therefore, 
x=l  in  this  equation,  which  corresponds  to  — 1  in  the 
given  equation. 

Therefore,  (1)  has  two  roots,  x=\  and  x= — 1:  hence, 
that  equation  is  divisible  x2 — 1. 

The  quotient  is,         4#2-|-3a; — 4=0, 

Whence  x=~3±z'J13 

8 

8.  x*+24x3—  2z2— 24z-fl=0. 

This,  like  the  preceding,  and  for  the  same  reason,  is 
divisible  by  x2 — 1. 

The  quotient  is,         #2-f24.r — 1=0, 
Whence  *==■— IZdzjUB. 

9.  x*—2x*—7x2—Sx+\6=0.  (1) 
Here  the  sum  of  the  coefficients  is  zero.     Hence,  a?=l 

for  one  root,  and  the  equation  can  be  depressed  to 

x3— x2— 8z— 16=0.  (2) 

Assume     x=nP.     Then 

ns  p*—ri>  p3_8wp_!  6==0 

Letrc=4;    then     64P3— 16P2— 32P— 16=0 
Divide  by  16,  4P3  —  P2—  2P— 1=0. 


82  KEY  TO  ALGEBRA. 

Here  the  sum  of  the  coefficients  is  zero.  Therefore, 
P=l.  But  x—?iP,  and  n=4,  P=\.  Whence,  rr=4,  for 
another  root  of  the  equation. 

Now  (2),  divided  by  x — 4,  produces 
ar2-j-3:r-J-4=:0, 

"Whence  x=— — -V  Imaginary. 

Another  Solution: 

The  attempt  to  extract  square  root  results  as  follows: 
#4_2z3—  lx2— 8z-j-16  (x2— x 
x* 

^2x^~—7x2 
2x2—x      —  2x3-\-  x2 

— 8x2— Sx 
If  the  remainder  were  -\-8x2  in  place  of  — 8#a,  this  ex- 
pression would  be  a  square. 

It  will  be  8x2,  if  we  add  -{-\6x2  to  each  member.  Then 
we  shall  have 

xx—  2x3-\-x2-\-8x2— 8x+16=\6x2, 
Or  .  .  .  (x2—  x)2+8(x2  —  x)+\6  =  16x2, 

Square  root  x2 — x-\-4=zh4x. 

10.  x*+2x3—3x2—4z+4--=0.  (1) 
Here  one  value  of  #  is  1.     Dividing  by  x — 1,  we  obtain 

x3+3x2  —  4=0. 
Here,  again,  x=l;  and  another  division  produces 

Square  root   .    .    (#+2)   (z+2)=0,     x=— 2     z=— 2. 

11.  x*—  2x3— 25.c2+26.r-f-120=0.  (1) 
Assume  x=nP.     Then 

niP<—2n3P3—  25w2P2-f2GttP+ 120=0 
p4_2P3_25yj2+26  p+i20===()! 
n        n2  n3  nl 

Letw=3.     Then   P^—fpz  —  ^pz+^P+^^O, 
Or  27P4  —  18/J3— 75P2+26P+40=0. 

Here  the  sum  of  the  coefficients  is  zero;  therefore,  P=\. 
But  n=o,  whence  x=3. 


EQUATIONS.  83 

If  in  (3)  we  assume  n—5,  the  coefficients  will  again  be 
zero,  showing  that  another  root  is  5. 

Now,  equation   (1)   divided  by  x — 3  or  x — 5,  or  their 
product,  will   produce  a  quadratic  whicb^fl^gi\]e  r  towp\ 
other  roots,  #= — 2  and  x= — 4.  Jy^^     of  the  ^ 

12.  r4  —  2x3+2x2— x^ofm  N I V  E  EjCSDX  T  Y 

(x3 — \)x=2x2(x — 1V^     -  ^ 

As  x  is  a  common  factor,  x=Q;    an3^ft^Jl  7^(^60^ ■»**• 
common   factor,  therefore'  x=\.     Dividing 
tors,  we  obtain         x2-]~x-\-l  =  2x, 

Or  x*—x=—\.    Whence  x^^^"3 

2 

1 3.  x4  —4x  3-\-8x2  — 32a-=0.  ( 1 ) 

Or (x— 4)x3+{x— 4)8#=0,     Whence    z=0, 

Or   .    .     x=4,     or      #2-|-#=0,     or    x=dtzj — 8. 

14.  x3  +5x2  +3a?— 9=0.  (1) 
Here  the  sum  of   the  coefficients   is  zero;    therefore, 

x=l.     And  by  division  we  obtain 

£2_|_6:r+l=0,         Whence     #=—3^78. 

15.  o;3_|_6,r3_  7x_  60=0.  (1) 
Assume  x=nP.     Then 

n3P3+6naP2—7nP—60=0. 

Suppose  n=S.     Then 

3)27P3+54P2— 21P— 60=0 
9P3-{-\QP2—  IP—  20=0. 

Here  the  sum  of  the  coefficients  is  zero;  therefore, 
x=S,  for  one  of  the  roots  o^pthe  equation.  The  other 
roots  are  — 4  and  — 5. 

16.  ^3-[-8^2  +  17^4-10=0.  (1) 
Change  the  2d  and  each  alternate  sign,  then 

x3—8x2  _f_17#— 10=0. 
Here  the  sum  of  the  coefficients  is  zero.     Hence,  x=\ 
for  the  last  equation;    or,  x= — 1  for  the  given  equation. 
Now  divide  the  given  equation  by  *-|-l,  and  the   quotient 
will  be  x24-7x-\-\0=0,         x=—2     or     —5. 


84  KEY  TO  ALGEBRA. 


17.                     x9—- 29x2 

-f  198z— 360=0. 

Place  x=?iP.     Then 

p3           4,V  [          . 

198  P     360_Q 

JT             —    | 

n 

n2          n* 

Assume  w=3.     Then 

(') 


p*  _?£p2  J_22P  —12=0 

3  3 

3/?3_29Jp2^_66p_40=0 

Here  the  sum  of  the  coefficients  is  0.     Therefore,  x=3; 
and,  dividing  the  equation  by  (x — 3),  we  obtain 
x2—  26*+ 120=0, 

x=6         or     20. 

18.  4x3  — 112z2  +  109ar—  27=0.  (1) 

Here  the  absolute  term  — 27  is  not  numerically  large 
enough  to  make  the  sum  of  the  coefficients  zero;   there- 
fore, we  take  an  operation  which  will  increase  it: 
p 
Place     x= —     Then 


2 

4P3      112P2  ,    109P 


•27=0 


8  4        '       2 

Multiply  by  2,  and     P3— 56P2+109P—  54=0.  (2) 

Here  the  sum  of  the   coefficients  is  zero.     Therefore, 
P=\.     Whence  x=\. 

Dividing  (2)  by  P — 1,  we  obtain 

P2—55P4-54==0. 
Here,  again,  P—\;  and,  of  course,  another  root  of  the 
original  equation  is  \,  and  it  has  two  equal  roots. 
Dividing  (1)  by  x — \,  a#d  we  obtain 
4x2— 110^+54=0, 
Whence   ....  #=27 


ON  INDETERMINATE  EQUATIONS. 


For  the  complete  solution  of  a  problem,  we  must  have  as 
many  independent  equations  as  unknown  quantities  to  be 
determined. 

When  this  is  not  the  case,  the  problem  is  indeterminate. 
For  example,  x-\-y—20.  x  may  be  one  or  two  or  three,  or 
5,  or  any  other  number,  whole  or  fractional,  under  20,  and 
y  will  take  the  remaining  part  of  20,  and  the  equation  is 
indeterminate  in  the  strictest  sense  of  the  term. 

If,  however,  we  restrict  the  values  of  x  and  y,  to  whole  or 
integer  numbers  in  the  equation  rc-f-?/=20,  x  cannot  have 
more  than  20 different  values,  when,  without  this  restriction 
x  might  take  an  infinite  number  of  values,  and  still  preserve 
the  equation  x-\-i/=20. 

In  some  cases,  the  number  of  answers  to  an  equation 
may  be  infinite,  and  the  particular  values  restricted  to  inte- 
gers.    The  following  is  a  general  case  of  the  kind 

ax — by—c.      This  givi  s  x=  —£— :  ,  where  y  may  be  any 

a 

value  whatever,  that  will  give     '       ,  a  whole  number,  but 

Cb 

numberless  such  values  of  y  can  be  found,  as  c-\-by  is  a 
magnitude  that  can  rise  higher  and  higher,  without  limit, 
according  to  the  assumed  value  of  y. 


86  KEY  TO  ALGEBRA. 

But  take  y  what  we  will,  and  the  equation  still  exists, 
and  therefore,  the  member  of  ansivers  for  x  is  unlimited  or 
infinite.  In  such  equations,  if  the  least  values  of  x  and  y 
are  required,  we  have  definite  problems. 

In  equations  like  the  following,  ax-\-by—c,  the  number  of 
answers  in  integer  numbers,  may  be  very  limited,  may  be 
only  one,  or   may  be  impossible.      The  equation  gives 

a 
Now,  if  c  is  very  large,  and  b  and  a  small,  y  may  take 
many  different  values  before  c — by  is  so  small  that  we  ean- 
not  divide  it  by  a,  and  obtain  an  integer  quotient. 

When  c  is  not  large  in  reference  to  b  and  &,  we  may  ob- 
tain only  one  value  of  y  and  #,  and  if  by  making  y=l,  we 

c 0 

find  — -,  a  proper  fraction,  the  problem  is  impossible. 
a 


3x-\-5y=\3.      a?=  '  Q  °y .      If  we  take  y—l.     x=-^, 

not  an  integer,  therefore,  y  must  not  be  taken  equal  to  one. 
Take  y=2,  then  a?=l,  both  integer  values,  and  the  only 
integer  values  that  will  answer  the  conditions  of  the  equa- 
tion. 3#-)-5^=6,  is  an  equation  in  which  it  is  impossible 
to  give  integer  values  to  both  x  and  y,  because  3— 1-5  is  more 

c—b .  ,       . 

than  6,  or  a-\-b  greater  than  c  or is  a  proper  fraction. 

The  equation  az-\-b?/=c.  is  always  possible  in  integers, 
when  c  is  greater  than  (ab — a — b).  and  a  and  b  prime  to 
each  other.  The  equation  is  sometimes  possible,  when  c  is 
not  greater  than  {ab — a — h.) 

The  equation  lx-\-l3i/==7\,  is  impossible  in  integers,  for 
both  x  and  y.  But  the  equation  lx-\-\3y=27.  is  possible, 
a?  =2  and  y=l.  Here  a  and  b  are  the  same  in  both 
equations.  ,  In  the    first  equation,  e=71 ;    c  is 

not  large  enough  to  make  the  equation  possible,  for  c,  71,  ia 


y 


INDETERMINATE' EQUATIONS.  87 

not  greater  than  (7*13 — 20),  but  if  c  was  any  number  greater 
than  71,  the  equation  would  be  possible -in  integers,  and  it 
is  possible,  with  some  numbers  less  than  71. 

If  a  and  b  are  not  prime  to  each  other  in  the  equation 
ax-^-by=c,  c  must  be  divisible  by  the  same  number  that  di- 
vides a  and  b,  or  the  equation  is  impossible  in  integers  for  x 
and  y. 

Examples. 

1.  Given  6a-f9y=32.     Or,  2a?-f3//=y. 

But  if  x  is  a  whole  number,  2x  will  be  a  whole  number, 
and  by  the  same  considerations,  3y  must  be  a  whole  number, 
and  two  or  more  whole  numbers  added  together  can  never 
make  a  fraction,  therefore  the  equation  2x-^-3y=  3f ,  or 
6r-|~9?/=32,  is  impossible  in  integers. 

In  cases  where  solutions  are  possible,  our  rules  of  opera- 
tion rest  entirely  on  these  considerations: 

1st.  A  whole  number  added  to  a  whole  number,  the  sum  is 
a  whole  number. 

2d.  A  whole  number  taken  from   a  whole   number,  the 
remainder  is  a  whole,  number. 
1        3d.   Multiply  a  whole  number  by  a  whole  number,  and  the 
■  product  is  a  whole  number. 

For  instance,  if  x  is  a  whole  number,  2.r,  3a?,  4x,  or  any 
integral  number  of  x  is  a  whole  number. 

2.  Given  3a?-f»5y=35,  to  find    x  and   y  in   whole  num- 

**»■     i—  |IW  r        *  "" 

bers.     a?=      Q  But  as  x  is  a  whole  number,  its  equal, 

o 

35— 5» 

or  — q~^i  must  equal  some  whole  number. 

Now  11 — y  being   a  whole  number  take  it  away,  and 

*  the  remainder   — -—  ,  must  also  be  a  whole  number. 
o 


88  KEY  TO  ALGEBRA. 

Sv 

But  y  being  a  whole  number,   --      is   a  whole  number, 

3 

ml  3y  ,  2—2;/     ?/4-2  .    . 

Therefore,  -  -j — -= — =— ^—   is  a  whole  number,  which 

number  call  ;?.     And  - '-  ==p.     Or,  y=3p — 2. 

For  the  least  value  of  y  make  p=l,  and  ?/  will  equal  1, 

35 5„ 

anda?= — = — =10.      Make  p=2,  then   ?/=4,  and  #=5. 
3 

Make  ^=3,  then  ?/=7,  and  a?=0. 

Hence,  ?/=l  or  4,  and  z=10  or  5,  are  the  only  results 
this  equation  admits  of 

In  operating  on  the  fractional  expression  — - — *,  it  was 

3 
our  object  to  work  down  the  coefficient  of  y  to  1 .  To  accom- 
plish ti>is  object,  we  cast  out  whole  numbers,  add  and  sub- 
tract whole  numbers  in  the  shape  of  fractions,  &c,  only- 
taking  care  to  keep  expressions  that  are  equal  to  integers, 
until  the  coefficient  of  y  becomes  one. 

3.  Given  35a? — 24?/=  68,  to  find  the  least  values  of  #  and 
y  m  integers. 

The  number  of  values    in  this  case  is  unlimited. 
_68+24?/_       33+24?/ 
*-      35      -1+~35" 

Hence,  — ~J~- — -  =  some  whole  number  ;  but  -~  is  also 
3d  3d 

Soy     33-4-24//     Ut/ — 33 

a  whole  number,     -p- '  -  •  =  -^    — =an  integer. 

3d  3d  3d  ° 

33j/-99_33//-29 

Ur'  ""35      ~      35~ 

35?/    33?/— 29     27/-J-29  _  .■        . 

-h^ qcl~ =     qc      —  a  wnoIe  number,  multiply  by  18, 

3d  3d  3d 

.  36?/+18-29         ,  % .  ,  y+32  .    . 

and  —    '        ■ — =3/-f-14-f--  - '   -  =  a  whole  number. 
3d  3d 

Therefore  V—^-=p.     Or,  y=Sbp— 32. 


INDETERMINATE  EQUATIONS  gg 

Take  p=l  for  the  least  value  of  y.  and  y—3.  Therefore 
*=4. 

4.  A  man  wishes  to  lay  out  $500  for  cows  and  sheep: 
the  cows  at  17  dollars  per  head,  and  the  sheep  at  2  dollars. 
How  many  of  each  did  he  purchase  1 

Let  z==  the  number  of  cows,  and  y  the  number  of  sheep. 
Then  17#-|-2?/=500.  We  know  this  equation  is  restricted 
to  whole  numbers ;  because  the  man  could  not  have  part  of 
a  cow,  or  part  of  a  sheep. 

To  find  the  least  number  of  cows,  transpose  17a?. 

Then  y=250— 8*— ^. 

*  2 

Now  as  y  must  be  a  whole  number,  and  250 — Sx  must 

x 
also  equal  some  whole  number,       must  be  a  whole  number. 

That  is,  the  number  of  cows  must  be  an  even  number ;  be- 
cause the  number  must   be  divisible  by  2. 

x 
Hence,  -^=P-     Or,  x=2p.     Makep=l,  and  a?=2,  the 

least  number  of  cows.     Then  #=233,  the  corresponding 
number  of  sheep. 

Now  if  the  man  wished  to  purchase  as  few  sheep,  and  as 
many  cows  as  possible,  we  should  transpose  the  other  term, 

500— 2y    on  ,    7—2?/ 
thus: *=~  17      =     +  ~T7    ' 

7 2?/ 

Therefore,  — =— --  =  a  whole  number.      Multiply  by  8, 

and  — — — -  =  a  whole  number,  to  which  add  — -±-  and  we 
17  17 

have    — ~^=3-| — ^.    Drop  off  the  whole  number  3,  then 

5     I      y 

-J^L=zp.     Or,  y=l7p — 5.      Making  p=l,  gives  #=12, 

the  smallest  number  of  sheep.     This  gives  #=28,  the  cor- 
responding number  of  cows. 

The  number  of  cows  or  a?,  may  be  any  one  of  the  even 
numbers  from  2  to  28.  a 


90  KEY  TO  ALGEBRA. 

5.  A  man  wished  to  spend  100  dollars  in  cows,  sheep,  and 
geese  ;  cows  at  10  dollars  a  piece,  sheep  at  2  dollars,  and 
geese  at  25cts.,  and  the  aggregate  number  of  animals  to  be 
100.     How  many  must  he  purchase  of  each  ? 

Let  x=.  the  number  of  cows,  y  the  sheep,  and  z  the  geese. 

Then  -  -  -  1  Qx+2y+ |  =  100.  (1) 
And  -  -  -  -  x-\-  ?/-|-*  =100.  (2) 
Clear  equation  (1)  of  fractions,  and 

40*+8#-r-;?=40G. 
,T_j_  y_j_2r==l00. 

39x+7y       =300. 

300—7?/     „  ,  27—7?/  ,    ,  . 

xz=  —        ''— H ^-1=a  whole  number. 

Or,  5(       S9     j^=  — 3^— -  =  a  whole  number,  add  -^ 

,  4y+135       40* -f- 1350  0.      ?/+24  ,    . 

and   Kg- ,  or  ^^—  =  ?/+34+  tjjj-  =  a  whole 

number. 

Therefore,  8tt-~  ^     Or,  ?/=39^— 24=15. 

This  value  of  y,  gives  a?=5.     Hence,  z=80. 

If  we  take  p=2,  we  shall  have  y=54;  then  a?  will  come 
a  minus  quantity  an  inadmissible  circumstance  in  any  prob- 
lem like  this.  Therefore,  5  cows,  15  sheep,  and  80  geese, 
is  the  only  solution. 

6  A  person  spent  28  shillings  in  clucks  and  geese;  for  the 
geese  he  paid  4s.  4d.  a  piece,  and  for  the  ducks,  2s.  6d.  a 
piece.     What  number  had  he  of  each? 

Let  #=the  number  of  geese,  and  ?/  the  number  of  ducks. 

Then.  52.r-f30.?/=28-i 2.     Or,  26*4-15?/= 168. 

We  will  now  show  another  operation  to  reduce  this  kind 
of  equations. 

Take  the  lowest  coefficient,  (in  this  example  it  is  15,)  and 
observe  whether  it  will  divide  the  other  numbers  or  not.    If 


INDETERMINATE  EQUATIONS.  yi 

it  will  divide,  reserve  the  whole  numbers,  and  omit  the  frac- 
tions. In  this  case,  15  will  not  divide  26,  and  will  divide 
168.     The  quotient  will  be  11,  disregard  the  remainder. 

Now  assume  p=x-\-y — 11.  Then,  since  x  and  y  are 
whole  numbers,  p  must  be  a  whole  number.  Multiply  by 
15,  and  transpose,  and  we  have 

15j9 — 15a; — 15y= — 165. 
2llr+15?/=     168. 

By  addition,    15;>{-lla?  =3.         Assume  q=x-]-p. 

Then  -  -  -  -  l\q— Up— Ux=0. 

Sum 11<H~  ^P  =«*•       Assume  r=p-\-<Zq- 

Or, 4r—  4p—  8q=0. 

Sum 4r  -(-  3q='3.     Assume  s=r-\-q — 1 

Or, 3s—  3/—  3?=—  3. 

Sum, 3s+     r  =0~ ~Or,  r=—3s. 

Now  having  worked  down  to  unity  for  a  coefficient,  the 
problem  is  essentially  reduced.  We  can  make  5=0,  hence 
r=0.  Then  in  the  last  assumed  equation  q=l,  and  in  the 
equation,  r=p-^-2q,  gives  p= — 2  ;  and  in  the  preceding 
assumed  equation,  q=x-\-p,  that  is  #=3,  and  the  first  as- 
sumed equation  gives  y=Q. 

Hence  3  geese  and  6  ducks  is  the  answer,  and  no  other 
numbers  will  do. 

7.  Divide  the  number   100  into  two  such  parts,  that  one 
of  them  may  be  divisible  by  7,  the  other  by  11. 
Let  7x=  one  part,  and  ll^=the  other. 

Then  7a;-|-lly=100,  and  x  and  y  must  be  whole  num- 
bers. Assume  p=x-\-y — 14.         (1) 

Then 7p—7x—  ly=— 98. 

But 7x+Uy=   100. 

By  addition,    -  -  7p         +4^=2 

Assume  q=p-\-y.        (2) 

Then 4q — ip —  4y=^0. 

Add,  and  -  -  -  -  4?        4.  3^=2. 


92  KEY  TO  ALGEBRA. 

4?+  3p=2. 

Assume  r—q^-p.         (3) 
Then 3r— 3q—  3p=0. 

By  addition  -  -  -  3r-\-  q  =&"      Or,  q=2—3r. 

Take  r=0,  then  <7=2,  and  p= — 2.  And  from  equation 
(2),  2= — 2-|-#,  or  #=4,  and  lly=44,  one  of  the  numbers, 
and  of  course  56  is  the  other, 

8.  Find  a  number  which  being  divided  by  6,  shall  leave 
the  remainder  2,  and  the  same  number  divided  by  13  shall 
leave  the  remainder  3. 

Consider  that  in  division,  the  divisor  and  quotient  multi- 
plied together,  and  the  remainder  added,  gives  the  number 
divided. 

Let  N  represent  the  number  divided,  x  and  y  the  quotients. 

Then  6*4-2= N,  and  13y+3=N. 

Consequently,  6a? — 13//=1,  an  equation  in  which  x  andi/ 
must  be  whole  numbers,  because  they  represent  the  whole 
numbers  of  the  division. 

Assume  p=x — 2y.  We  take  2y  because  6  is  contained 
in  13,  twice. 

Then 6p— 6*4-12?/=0. 

And     ......  6o? — 13?/==1. 

Add,  and  -  «  -  -  6p  — //=1.     Or,  y=6p — 1. 

For  the  smallest  value  of  y  we  must  take  p=\.  Then 
7=5,  and  13?/-|-3=68,  the  answer. 

9.  What  number  is  that  which  being  divided  by  1 1,  leaves 
d  remainder  of  3,  divided  by  19,  leaves  a  remainder  of  5, 
and  divided  by  29,  shall  leave  a  remainder  of  10. 

Let  N  be  the  required  number,  and  x,  y,  and  z  the  several 
quotients,  and  of  course  they  must  be  whole  numbers. 
Then  1 1*4-3= N,  and  \9y-\-  5=N,  and  29;:4-10=N. 

Hence,  *J?^t?,  and  *J2t&.    19^=29*4-5. 


INDETERMINATE  EQUATIONS.  93 

Or, 19y— 29*=5. 

Assume  p=y — %.         (1) 
Then \9p— \9y+\ 9z=0. 

Add,  and   -  -    19^  — 10z=5. 

Assume  q=p — z.         (2) 
Then    -  -  -  -    10g—  10p+10z=0, 

By  addition  -    lO^-j-  9p  =5. 

Assume  r=q-\-p.         (3) 
Then  -   -   -   -     9r—  9q—  9p=0. 

Add,  and    -  -  -  9r+     ?  =5.     Or,  ?=5— 9r. 

By  returning  to  equations  (3)  and  (2),  we  find  p=\Or — 5, 
and   z=19r — 10.      Not  only  must  z  be  a  whole  number, 

but  to  make  x  a  whole  number,  -  "£-  must  be  a  whole  num- 
ber.    Substituting  the  value  of  z  in  this  last  expression,  and 

.         551r— 283        ,    .  .  CA      oc  ,  r— 8 

we  nave  ■ — a  whole  number,  or  50r — 25-| — — —  a 

whole  number.     Therefore,  — — -  must  be  a  whole  number, 

which  call  i\  then  r = 1 1  *-|-8.  Let  *=0,  then  r=8,  z=z 1 42, 
and  29z-f-10=N=4148,  the  number. 

10.  Required  the  least  number  that  can  be  divided  by  each 
of  the  nine  digits  without  remainders. 

Let  x—  the  number. 

_,.         xxxxxxxx  „   . 

The"   2'    3'    V    5'    6'    7'    8'    9'  mUSt  aU  be 

whole  numbers. 

Now  if  we  make  -  a  whole   number   ~j   and  — ,    the 

double,  and  quadruple  will  be  whole  numbers  of  course. 

x  .  x 

Also  if      is  a  whole  number,       will  be  a  whole  number, 
y  o 

Therefore  we  have  only  to  find  such  a  value  of  x  as  will 

flf*  fp  ffi  ff*  ft*  f 

make  -,    5,    -j    — ,    -   whole  numbers.    — ,  may  also 

y     8     7     o     5  0 

be  cast  out,  on  consideration  that  6=2-3,  and  2  3  are  factors, 
one  cf  9,  the  other  of  8,  in  preceding  expressions. 


94  KEY  TO  ALGEBRA. 

X  X         X         X 

Hence  we  have  only  to  make  -  ,    — ,    — ,    — ,  whole 

9      8      7      5 
numbers.     Put  x=9p. 

Then   ^=jp_[_^.     Hence,  p=8q.     Then  z=9p=72?. 

— -=  a  whole  number.     -    =  a  whole  number,  or  -=a4- 
7  7  7      *  ' 

.     Make  -J[  =r,  or  <?=7r.      In  the  same  way  we  find 

r=5s.     Take  5=1,  then  r=5.    ?=35.     a?= 72-35 =2520. 

N.  B.  By  the  aid  of  the  Indeterminate  analysis,  combined 
with  some  well  known  properties  of  numbers,  we  may  some- 
times solve  miscellaneous  problems  in  a  very  summary 
manner.     For  example,  we  refer  to  the  following. 

11.  The  product  of  five  numbers  in  arithmetical  progres- 
sion is  10395,  and  their  sum  is  35.    What  are  the  numbers  1 

As  the  number  of  terms  is  odd,  let  x  represent  the  mid- 
dle term,  and  y  the  common  difference.  Then  (x — 2y), 
(x — y),  x,  (x-\-y),  (x-\-2y),  will  be  the  numbers,  and  their 
sum  5#=35,  or  x=7. 

Now  as  10395  is  the  product  of  all  the  numbers,  7  must 
be  one  of  its  factors.  Therefore  divide  by  7,  and  we  have 
1485  for  the  product  of  the  four  remaining  terms,  but  as 
this  number  ends  in  5,  it  can  be  divided  by  5,  and  it  is  more 
than  probable  that  5  is  one  of  the  numbers,  and  the  one 
preceding  7,  therefore,  2  is  the  common  difference,  and  3, 
5,  7,  9,  and  11,  the  numbers. 

12.  Given  the  sum  of  the  squares  of  three  numbers  =195, 
the  sum  of  their  cubes  =1799,  and  their  continued  product 
=385,  to  find  the  numbers. 

By  the  common  rules,  without  considering  any  circum- 
stances, this  problem  would  produce  equations  of  high  order, 
and  difficult  of  solution.  But  let  us  call  to  mind  the  fact, 
that  the  numbers  cannot  be  fractional,  if  they  were,  their 
squares,  cubes,  and  product  would  not  probably  come  whole 
numbers.     Also,  some  of  the  numbers  must  be  under  10; 


INDETERMINATE   EQUATIONS.  95 

if  even  two  of  them  were  over  10,  the  cubes  and  product 
must  be  larger  than  the  numbers  mentioned.  From  consid- 
erations like  these,  we  decide  that  the  answer  must  be  whole 
numbers,  and  some  of  them  under  10.  Noav  as  the  product 
of  the  three  is  385,  and  the  sum  of  their  squares  is  195, 
both  numbers  ending  in  5,  it  is  so  probable  that  one  of  the 
numbers  is  5,  that  we  shall  so  consider  it,  and  let  x  and  y 
be  the  other  two  numbers,  then  a:2 -|-?/2 -{-25 =195,  and  hxy 
=385.  Equations  from  which  we  readily  find  one  number 
to  be  7,  the  other  11. 

Now  by  trial  we  find  53+7~+lP=1799,  and  therefore 
these  are  in  fact  the  numbers. 


Section  XX. 


To  determine  the   number  of  solutions  an  equation 
may  admit. 

It  has  already  been  observed  that  an  equation  in  the  form 
of  ax — by=c  admits  of  an  infinite  number  of  solutions, 

as  x=     '    *   a  quantity  all  positive,  and  the  only  restriction  is 

to  assume  y  of  such  a  value  that  the  numerator  may  be 
divided  by  a. 

But  equations  in  the  form  of  ax-\-by=c.  Then  x= - 

and  the  numbers  of  solutions  depends  on  the  relative  values 
of  c,  5,  and  a. 

If  c  be  very  large  in  relation  to  b  and  a,  as  we  have 
before  observed,  the  equation  may  have  many  solutions, 
otherwise  not. 

We  come  now  in  a  general  manner  to  determine  the  num 
ber  of  solutions  an  equation  of  this  form  may  have. 

Let  ax-\-by-=c.    Assume  ax! — by'=l.     Which  equation 


96  KEY  TO   ALGEBRA. 

is  always  possible,  and  from  which  x'  and  y'  can  be  known 
in  integers. 

Multiply  the  assumed  equation  by  c,  and   acx' — bcy'=c. 
Put  the  two  values  of  c  equal  to  each  other,  and 
ax-\-by=acx' — bey' 
x=cx' — b(y-\-cy').     Or,  x=cx' — bm.         (1) 

y=;a(C^j^)— cy'.     Or,  y=am—cy'.       (2) 

In  these  theoretical  equations,  (1)  and  (2),  m  has  different 
values,  it  being  an  arbitrary  number  taken  at  pleasure,  so 
that  ex'  may  be  greater  than  bm,  and  am  greater  than  cy'  to 
render  x  and  y  positive. 

But  if  no  such  value  of  m  can  be  found,  it  is  proof  that 

values  of  x  and  y  do  not  exist  in   positive  integers,  and  on 

the  contrary  as  many  suitable  values  of  m  as  can  be  found, 

so  many  solutions  will  the  equation   admit  of,  and  no  more. 

Now  as  -  -  -  mb  <^  ex.  and  am  ^>  cy' 

r\   ■  ^cx'         A  -^   CV' 

Or, m<—-.  and    m  >  — 

^  b  '  ^    a 

That  is  m  at  the  same  time  is  found  to  be  greater  and  less 
than  known  quantities,  therefore  its  limit  or  range  is  found. 

For  instance,  if  m  must  be  greater  than  30,  and  less  than 
40,  we  conclude  that  it  may  be  any  number  between  30  and 
40,  and  the  number  o£  different  values  it  can  take  is  9. 

We  perceive  that  the  difference  between  the  integral  parts 

of  — —  and  —  will  express  the  range  of  m,  and  the  number 

of  different  solutions  which  the  equation  admits  of,  (except 
in  certain  cases ; )  as  m  is  more  than  one  of  these  fractions 
and  less  than  the  other,  the  difference  between  the  expressions 

—   and  —  is  sometimes  one  more  than  the  number  of  dif- 
o  a 

ex 
ferent  values  of  m,  such  is  the  case  when  —   is  an  integer, 

in  such  cases,  subtract  one  from  the  difference  of  these  quan- 
tities for  the  range  of  m,  but  this  case  very  seldom  occurs. 


INDETERMINATE  EQUATIONS.  97 

N.  B.  In  making  use  of  the  expression  —  and  —  care 

0  '  Cb 

must  be  taken,  not  to  take  their  difference  as  factional  ex- 
pressions, their  absolute  difference  is  not  wanted,  it  is  the 

ex'        ciif 
difference  between  the  integral  farts  of  —  and-1-. 

°       r  b  a 

Example. 
Required  the  number  of  integral  solutions  to  the  equation 

7*4-%=  100. 
Find  the  least  value  to  a?',  y'  in  the  equation^' — 9/=l. 

lx' — 9y'=l.     Assume  p— x' — y' 
Then  -  -  lp— 7x'+7y'=0. 


Add,  and   lp         — 2y'=l.     Assume  q=3p — y' 
Then  -  -  2q— 6p+2i/=0. 

Add,  and  2q-\-  p=l.     Or,  p=l — 2q. 

Take  £=0,  then  p=l.     y'—%     #  =4. 

mL       ex'      100-4       j  cy       100-3 
Then  — =— — -  and  -£  =  — —  . 
b  9  a  7 

That  is  ^=441  and  ^-=42« 
b  9  a  1 

Disregarding  the  fractions,  the  difference  of  the  integral 

parts  is  2,  that  is  there  are  two  integral  solutions  to  the 

equation. 

400     300 

9         7 

....                  2800    2700     100     13_ 
fractional  form,  thus:   — ^ .— -   =—=137. 

Here  the  integral  difference  is  one,  which  without  this 
caution  might  be  taken  for  the  number  of  solutions.  The 
integral  difference  in  this  case  is  not  the  difference  of  the 
integrals. 

Observe  in  this  example  44£  and  42|,  the  fractional  part 

of  —  is  |,  and  the  fractional  part  of  —  is  f ,  the  former  is 
o  a 

less  than  the  latter*  in  such  cases  the  integral  parts  must  be 
taken  separately. 


If  we  had  taken  the  difference  between  — — ,   in  a 


98  KEY  TO  ALGEBRA. 


ex 
But  when  the  fractional  part  of  — -  is   not  less  than  the 

b 

cv 
fractional  part  of  -— ,  but  equal  or   greater  than  it,  we  may- 
find  the  number  of  solutions  by  taking  the  difference  of  the 

ex'     cy'      _    , 
expressions  — .     Reduce  to  a  common  denominator, 

and  take  the  difference  of  the  numerators,  and  we  will  have 

— - — —  ;    but  ax' — by'=  1.      Therefore,  we  have  — 

ab  ab 

for  the  number  of  solutions,  at  once. 

Example. 

What  number  of  integral  solutions  will  the  equation 
1 9,z-f-l 3j/ =2000  admit  of?  Ans.  17. 

~9X13=117)  2000(17. 
But  as  we  cannot  know  whether   the  fractional  part  of 

C11  CI  J 

~y~  is  not  less  than  the  fractional  part  of  —    we  cannot  be 
b  r  a 

sure  that  dividing  c  by  ab  will  give  the  true  number  of  solu- 
tions.    It  either  will  be  the  true  number  or  one  less. 

The  equation  5#-|-9#=40  admits  of  no  solution  in  whole 
numbers,  c=40  will  not  be  divided  by  aZ>=45. 

Now  take  the  equation  hx' — 9#'=1. '  And  we  find  x'=2, 

and  y'=l. 

ex'     80  cy'     40 

Therefore,  — =  n  =8£,  and  — =  -  =8.    Now  as  there 
b       9        y  a       b 

is  no  difference  between  these  integral  parts,  it  indicates  as  it 
should,  that  there  is  no  solution. 

But  let  us  take  5#-f-9//=37,  the  same  equation,  except  a 
smaller  value  of  c.  If  c  would  not  divide  by  ab  before, 
much  less  will  it  now.     Yet  in  this  last   equation  we  have 

ex'     74 

a  solution,    a?' =2,  y'=l,  as  before,  and  c=37.  — =-^-=8|- 

and  —  =  _  =7|.     Here  the  difference  of  the  integrals  is  1, 
a       5 

and  indicates  one  solution.     x=2,  y=3. 


INDETERMINATE  EQUATIONS.  99 

How  many  solutions  will  the  equation  2a:-)-5^=40  admit 
of? 

The  auxiliary  equation  2a?' — 5/=l,  gives  #'=3,  y'=l. 

«*[_.  24.      ^=20.     Or,  4  solutions. 

C2?' 

But  observe  that  ~r  in   this  case,  is  a  complete  integral, 

24 ;  agreeable  then,  to  previous  considerations  we  must  de- 
duct one,  and  the  number  of  solutions  are  but  3,  as  follows: 
#=5.  10.  15.  #=6.  4.  2,  and  no  other  solution  can  be 
found. 

What  number  of  solutions. in  whole  numbers  can  be  found 
for  the  equation  3x-\-5y-\-lz=\00. 

As  x  and  y  each  cannot  be  less  than  one,  z  cannot  be 

greater  than = =131.    That  is,  z  cannot  be  greater 

than  13,  in  whole  numbers.  Now  suppose  2=1,  and  the 
equation  becomes  3x-\-oy=93. 

The  number  of  solutions  for  this  equation,  found  as  pre- 
viously directed  is  6.     rm,of-    J  a?  =  26.  21.  16.  11.    6.     1. 
inatis^?/==    3     6     9  ]2  15  1Q 

Now  x  and  y  can  make  these  6  changes,  and  z  be  con- 
stantly equal  to  1,  and  satisfy  the  primitive  equation. 

Take  z=2,  and  the  equation  becomes  3#-j-5<y=86. 

This  equation  has  also  6  solutions,  z  being  through  all  the 
changes  of  x  and  y  equal  to  2. 

Now  take  z=3,  then  the  original  equation  is  3z-|-5y=79. 
This  equation  has  five  solutions. 

Now  take  z=4,  then  3.r4-5y=72.  This  equation  has 
four  solutions. 

Take  z=5,  then  3#-|-5y=65.  This  equation  has  four 
solutions. 

Take  2=6,  then  3z-[-5y=58.  This  equation  has  four 
solutions. 

In  this  manner,  by  taking  z  equal  to  all  the  integers  up 
to  12  in  succession,  we  find  41  solutions. 


100  KEY  TO   ALGEbKA. 


THE  DIOPHANTINE  ANALYSIS. 

The  Diophantine  Analysis  teaches  hew  to  find  square 
and  cube  numbers  under  given  conditions,  or  having  given 
relations  to  each  other. 

Examples. 

Case  1st.  Find  such  a  value  of  x  as  will  make  the  ex- 
pression ax-\-b  a  square. 

Put  ax-\-b=n2,  n2  being  any  square,  it  is,  therefore,  an 
indefinite  problem.   ' 

_  .  .  n2 — b       .  - 

From  the  equation  x= ;  take  n  equal  to  any  num- 
ber whatever,  and  a  and  b  being  known,  x  becomes  known. 

Eight  times  a  certain  number  added  to  9,  makes  a  square. 
What  is  the  number  ? 

Let  a?  =  the  number.      Then   Sx-\-9=n2,  that  is  any 

o    i     n       a  n2—b     n2~9       A 

square.     a=o,  o=9,  and    x= = — — —  .      Assume  n 

a  o 

=7,  then  #=5,  the  required  number.  But  there  are  many 
other  numbers  that  will  answer  the  condition  according  as 
we  assume  n  more  or  less. 

Find  x,  such  that  the  following  expressions  shall  be  square 
numbers. 

9*4-9.     7#+2.     3x— 5.     a?+|. 

All  these  correspond  to  the  general  expression  ax-\-b. 

Case  2d.  Any  algebraic  expression  in  the  general  form 
of  ax2-\-bx,  may  be  made  a  square  by  supposing  its  square 
root  equal  px;  x  must  be  in  some  part  of  the  root,  because 
the  expression  contains  a:2,  that  is  some  function  of  a?.  Now 
if  px  is  the  root,  ax2 -\-bx=p2  x2 .     Divide  by*,  &q    and  we 

have  %=--£ —  •    p  may  be  any  assumed  value  whose  square 

is  greater  than  a, 


DIOPHANTINE   ANALYSIS.  101 

Examples. 

Six  times  the  square  of  a  certain  number,  added  to  five 
times  the  number  is  a  square.     What  is  the  number  ? 

5 

6x2 -\-5x=p2  x2 .      Or,  x= -• 

p2 — 6 

Here  it  is  obvious  that  p2  must  be  greater  than  6,  other- 
wise it  is  unlimited.     Take  p=3,  then  a?=  f  =  If. 

x2 
Find  the  value  of  x  to  make  —  -\-3x  a  square. 

Ans.  #=6. 

Case  3d.  Any  algebraic  expression  in  the  general  form 
of  x2±bx-\-c,  can  be  made  a  square,  by  putting  xdzp  equal 
its  square  root. 

We  can  if  we  please  take  x — p  for  the  root  in  all  such 
cases.  Then  if  p  is  less  than  x,  the  square  is  diminished,  if 
greater,  the  whole  root  will  be  essentially  minus,  but  the 
square  will  be  plus,  and  may  rise  to  any  amount.  Therefore 
x — p  is  far  more  general  than  x-\-p. 

Case  4th.  Any  expression  in  the  form  of  ax2  dzbx-\-c2 , 
can  be  made  a  square,  by  taking  its   root   equal  to  cdzpx. 

It  will  be  observed  that  x  must  be  in  the  root  of  the  pre- 
vious expression,  because  it  has  x2,  and  c  must  be  in  the  root 
of  this  last  expression,  because  it  contains  c2. 

In  the  first  we  have  x2±bx-\-c=x2±2px-{-p2.  Or, 
p2 — c 

In    the  second,  x2±bx-\- c2—c2±:  2cpx-{-p2x2.      Or, 

x±:b=  ±3cp-\-p2x.      Or,  x=  -     ^~-      In  both  cases  as- 

Idap2 

sume  p  of  any  convenient  value  to  render  x  positive,  and  as 

small  as  possible. 

Find  a  number   such,  that  if  it  be  increased  by  2  and  5 
separately,  the  product  of  the  sums  shall  be  a  square. 
Let  z=the  number,  then  (a?4-2)  (#+5) =a?2 +73+10, 


102  KEY   TO   ALGEBRA. 

must  be  a  square.  b=l,  c=10.   General  solution,  x=J—^ — 

2p±7 

Now  p2  must  be  more  than  10 ;  hence,  take  j?=4,  and 
*=T65  =  |,  the  least  number  that  will  answer  the  conditions. 

Case  5th.  An  expression  in  the  form  of  ax2 -\-bx -\-c, 
where  neither  the  first  nor  the  last  terms  of  the  expression 
are  squares,  neither  branch  of  the  root  can  be  directly  found, 
and  the  expression  cannot  be  made  a  square,  unless  we  can 
separate  it  into  two  rational  factors,  or  unless  we  can  first 
subtract  from  it  some  simple  binomial  square,  and  can  then 
divide  the  remainder  into  two  rational  factors. 

By  reminding  one  of  the  nature  of  quadratic  equations, 
all  may  perceive  that  the  expression  ax2-\-bx-\-c  must  be 
the  product  of  two  factors,  but  whether  rational  factors  or 
not  is  the  subject  of  inquiry. 

To  find  the  factors  which  make  the  product  ax2 -\-bx-\-c, 
put  this  expression  equal  to  0,  and  work  out  the  values  of  x 
thus,  ax2-\-bx-\-c=0.     Or,  ax2-\-bx= — c.     Complete  the 
square,  and    4a2 x2 -\-\abx-\-b2  =  b2 — \ac. 
Or,  2ax-\-b=  ±J  \b2—4ac.\ 

We  now  perceive  that  the  values  of  x  must  be  rational, 
provided  J  \b2 — 4ac\  is  a  complete  square.     If  it  be  so,  let 

la-J  \^-iac\-l=nh  and  -I  J  («•-*«}-  £•* 

Then  the  two  values  of  x  are  x  =  m  and  x=n,  and 
(a? — m)  (x — n),  are  the  factors  which  will  give  the  expression 
ax2  -\-bx-^-c. 

Examples. 

1.  Find  such  a  value  of  x  as  will  make  6a?2-|-13x-|-6  a 
square. 

Here  «=6,  £=13,  e=6.  £2=169,  4ac=144,  Z>2— lac 
=s25  and  J  \  b2—4ac }  =5.  Now  12o:+l3=±5.  *=--§ , 
Or,  *=— |.     Or,  3z+2=0,  and  2*4-3=0. 


DIOPHANTINE   ANALYSIS.  102 

That  is,  (3x-\-2)  (2#-|-3)  will  produce  the  expression 
6x2+13tf+6. 

Now  to  make  the  expression  a  square,  put 
(3*+2)  {2x+3)=p2{3x+2)2 

Then  2x+3=p2 (3z+2.)     And  »=-?£_. 

Take  p=l,  and  x=l.  We  might  have  seen  at  first, 
that  in  this  particular  expression,  the  value  of  x  being  1, 
would  make  it  a  square,  as  6-j-13-|-6=25,  a  square. 

That  is  in  all  cases  when  the  sum  of  the  coefficients 
make  a  square  number,  the  value  of  x  may  be  one. 

2.  Find  such  a  value  of  x  as  shall  render  the  expression 
13x2-\-]5x-\-7*a.  square. 

Here  as  neither  the  first  nor  last  terms  are  squares,  nor 
b2 — 4^  a  square,  the  expression  cannot  be  made  a  square, 
unless  we  can  separate  the  remainder  into  factors  after  taking 
away  some  simple  square.  But  in  this  case,  \ac  is  greater 
than  b2,  we  must  then,  in  subtracting  our  square,  diminish 
a  and  c  and  increase  b. 

To  accomplish  this  end  we  will  subtract  the  square  of 
* — 1,  not  x-\-\. 

That  is 13z2+15z+7. 

Subtract x2—  2.r+l. 

Remainder,-  -  -  -  l2x2-\-\lx-\-Q. 

In  this  last  expression,  a=12,   £=17,    cz=6.      Hence, 
b2 — 4ac=289 — 288=1,  a  square  ;  we  are  now  sure  ra- 
tional factors  can  be  found  to  produce  the  expression 
12z2+17a?-|-6. 

By  assuming  122-}-17a;-|-6=0,  and  finding  the  values  of 
s  by  the  quadratic,  (merely  to  get  the  factors,)  we  find 
«ass— |,  and  a?== — |.     Or,  3; +2=0,  and  4*+3=0. 

And  (3z+2)(4z+3)=12*2+17z+6.* 

♦The  values  of  x  used  to  obtain  these  factors  have  no  connection  with  tat 
values  of  z  to  render  the  original  expression  a  square. 


104:  KEY  T0  ALGEBRA. 

Now  the  original  expression  is  the  same  as  (a? — l)2-f» 
(3a?-f 2)  (4a?+3.)  Put  its  root  equal  to  (x—  l)+^(3a?+2)) 
and  square  this  root,  and 

(a?—  l)2+(3a?+2)  (4a?-f-3)=(a?— 1)2+  2^(a?— 1)  (3a?+  2)+ 
/(3a?+2)2.  By  reduction,  4a?+3  =  2p{x—l)+f[3x+2. 
Or,  (— 4+2p+3/)a?=2p+3— 2p2. 

Therefore    *-2^+3— 2^ 
inerelore,  ^^—-^ 

Take^=l,  then  a;=3,  and  13:r-J-15a?-}-7=169a  square, 
Ryan  makes  a;  in  this  example  equal  i  and  the  expression 

equal  *§ l  a  square,  but  an  integer  number  is  always  more 

satisfactory. 

3.  Find  such  a  value  of  a?  as  will  render  14a?2-j-5a? — 39, 
a  square. 

After  a  few  trials  this  expression  is  found  to  be  the  same 
as  (2a? — l)2-|-(5a? — 8)(2a?-|-5.)  Assuming  its  root  to  be 
2a? — l-\-p(5x — 8  )     Then  by  squaring  the  root,  making  it 

equal  to  its  power,  and  reducing,  we  find  a? = ,  7TA       </ 

Assuming  p=l,  a?=1T5,  and  the  expression  equals  36,  a 
square.  Other  values  can  be  found,  by  assuming  different 
values  to  p. 

4.  Find  such  a  value  of  x  as  shall  make  2a?*-(-21a?-|-28 
a  square. 

After  a  little  inspection  we  find  this  expression  equal  to 
(a;+4)2+(a?+l)  (a?+12.)     Now  if  we  make 

(z+4)2+(*+l)  (a?+12)=  {(*f4>-p(*+l.)J» 

After   reduction,  we  shall  find  a?=    ,     ^      1  • 

pr — 2/? — 1 

Assume  p=4,  then  a?=4,  and  the  original  expression  is 
144  a  square. 

If  (a?+4)2+(a?+l)(a?+12)={(a?+4)-^(a?+12)}2,  we 
shall  find  a?=  ^~  P~  .     If  we  take^=l,  a?=f .     If  we 


DIOPHANTINE   ANALYSIS.  105 

take  2?=!,  a?=8,  and  we  might  find  many  other  numbers 
that  would  answer  the  conditions  of  the  expression. 

Case  6th.  When  we  have  an  expression  in  the  form  of 
a^x^-^-bx^-^-cx'-^-dx-j-e,  we  can  assign  a  value  to  x  that  will 
make  the  whole  expression  a  square,  if  we  can  extract 
three  terms  of  its  root. 

Assume  such  terms  as  the  whole  root,  square  the  root  so 
assumed,  making  it  equal  to  the  given  expression,  and  by 
reducing,  we  shall  have  a  value  of  x  which  will  make  the 
original  expression  a  square. 

Example. 

Find  such  a  value  of  x  as  shall  make  4x4-\-4x3-\-4x2-\- 
2x — 6  a  square. 

We  commence  by  extracting  the  square  root  as  far  as 
three  terms,  and  find  them  to  be  2#2-|-#-|-£.  Now  assume 
4z  +4x3-\-4x-+2x— 6=(2z2+#+!.)2 

3a?      9 

Expanding  and  reducing,  we  have   2x — 6=— — |-— . 

And   x=\3\. 

Essentially  the  same  method  must  be  pursued  in  other 
problems  of  the  like  kind. 

Case  7th.  Find  such  a  value  of  x  as  shall  render  2ar2-f-2 
a  square. 

Expressions  of  this  kind,  when  neither  a  nor  c  are  squares, 
nor  b2 — \ac  a  square,  and  which  cannot  be  resolved  into 
factors,  presents  an  impossible  case,  unless  we  can  first  find 
by  inspection,  some  simple  value  of  x  that  will  answer  the 
condition. 

In  the  present  example,  it  is  obvious  that  if  #=1,  the 
expression  is  a  square.  But  we  wish  to  find  other  values  than 
1  that  will  render  this  expression  a  square,  and  having  found 
that  one  will  answer,  we  are  now  enabled  to  find  other  val- 
ues, thus : 

Let  a?=  l±y.     Then  x2=lzk2y+tf. 


106  KEY  TO   ALGEBRA. 

And  2x2-|-2=4d=%-{-2yz.  Here  the  original  expression 
is  transformed  into  another  expression,  having  a  square  for 
its  first  t^rm. 

Now  we  must  find  such  a  value  of  y  as  shall  make 
4-j-4?/-|-2y2  a  square. 

Assume   4-}-4^-|-2^2=  (2 — my)2— 4 — 4;my-\-m2yz.     Or, 

a   i  r>  /I       i     \  T1  4(m-f-l) 

4+~j/=  — lm-\-m-y.     Hence  #=s  -^f-  9  i  m  mav  De  any 

number  greater  than  one.  Put  m=2.  Then  #=6,  and 
5c=l-(— y=7,  and  the  original  expression  2;r2-f-2=98-}-2= 
100,  a  square. 

N.  B.  It  often  occurs  incidentally  in  the  solution  of  prob- 
lems, that  we  must  make  a  square  of  two  other  squares. 
This  can  be  done  thus :  Let  it  be  required  to  make  x2-\-y2 
a  square.     Assume  x=p2 — q2,  and  i/=2pq. 

Then, x2=p* — 2p2q2-\-q\ 

And, y'2=         4p*q3. 

Add,  and  -  -  x2-\-y-=p  -\-~p2q2-{-q4,  which  is  evidently 
a  square,  whatever  be  the  values  of  p  and  q.  We  can, 
therefore,  assume  p  and  q  at  pleasure,  provided  p  be  greater 
than  q. 


Double  and   Triple   Equalities. 

In  the  preceding  section  it  was  only  necessary  to  find  such 
a  value  of  the  unknown  term  as  to  render  a  single  expres- 
sion a  square.  But  there  are  problems  where  it  becomes 
requisite  to  find  such  a  value  of  the  unknown  term  as  to 
render  several  different  expressions  squares  at  the  same  time. 
And  this  is  called  double  and  triple  equalities. 

Case  1st.  As  a  general  equation  for  double  equality,  let 
it  be  required  to  find  such  a  value  of  x  that  ax-\-b  may  be 


D10PHANTINE   ANALYSIS.  107 

a  square,  and  the  same  value  of  x  give  cx-\-d  a  square. 

f2 Q 

Let  ax-{-b=t2.       Then  x. 


And  cx-^-d=p'1.       Then   x= 


a 
p2 — d 


Therefore  =  — .      Or,   ct2 — cb  =  ap2 — ad. 

a  c  $ 

Tranpose  cb  and  multiply  by  c  and  we  have 
c2t2-=acp2-\-c2d — acd. 

As  the  left  hand  side  of  this  equation  is  a  square,  what- 
ever  may  be  the  values  of  c  and  t,  it  is  now  only  requisite 
to  find  such  a  value  of  p2  as  shall  render  the  other  side  a 
square,  which  can  be  done  by  some  one  of  the  artifices  in 
the  preceding  section. 

To  illustrate  this  we  give  the  following  definite  problem. 

The  double  of  a  certain  number  increased  by  4,  makes 
a  square,  and  five  times  the  same  number   plus  one,  also 
makes  a  square.     What  is  the  number    * 
Let  x  represent  the  number. 

Then    2^+4=  *2] 


J>      From    which   < 
And      5a?-j-l=p2 


x=  — 


5 

Hence  5*2— 20=2p2—2.  Or,  5*2=2j92+18,  multiply 
by  5,  and  25^  =  10;?2+90. 

The  left  hand  side  of  this  last  equation  being  a  square, 
whatever  be  the  value  of  t,  it  is  now  only  necessary  to 
find  such  a  value  of  p2  as  to  make  10^2-f-90  a  square,  an 
expression  which  corresponds  to  case  7  of  the  last  section. 
We  therefore  cannot  proceed  unless  we  find  by  trial,  by 
observation,  by  intuition  as  it  were,  some  simple  value  of 
p  that  will  make  10/>2-|-90  a  square,  and  we  do  perceive 
that  if  p= 1,  the  expression  will  become  100,  a  square. 

Now  if  ;?=1,  it  will  give  a  definite  and  positive  value  to 
*,  and  the  problem  is  solved.  If  not  we  must  find  other 
values  of  p. 


108  KEY  TO   ALGEBRA. 

But  we  have  found  that  #=  — - — ,   and  if  p=l,  £=0, 

and  the  original  expressions  2z-|-4,  and  5#-|-l,  become  4 
and  1,  squares  it  is  true,  which  answer  the  technicalities,  but 
not  the  spirit  of  the  question. 

^To  find  another  value  of  p.  Put  p  =  1  -|-  q.  Then 
10^2+90  =  100*-f20;?+2?2.  To  make  this  a  square 
assume  100-\-%0q+2q2  =  {l0—  nq)2  =  100— %0nq-\-n2q2. 

By  reduction,   q=  -  —}—  •".  .     Now  n  must  be  so  taken, 

J  a       n2 — 10  7 

that  n2  is  more  than  10:  take  ti=5  and  ^=8,  ^=9,  then 
#=16  and  the  original  expressions  2#-f- 4  =  36,  a  square, 
and  5#-|-l=81j  a  square. 

Case  2d.  A  double  equality  in  the  form  of  ax2-\-bx=Q 
and   cx2-\-dx,  also  equals   a  square,  may  be  resolved  by 

making  *=— ,  tlten  the  expressions  will  become  —  (a-\-by) 

and  — -  (c-\-dy,)  which  must  be  made  squares. 

But  if  we  multiply  a  square  by  a  square,  or  divide  a 

square  by  a  square,  the  product  or  quotient  will  be  square. 

Now  as  each  of  the  preceding  expressions  are  to  be 

squares,  and  as  they  obviously  have  a  square  factor  —  it  is 

only  necessary  to  make  a-\-by,  and  c-^-dy  squares,  as  in  the 
first  case. 

We  may  also  take  another  course  and  assume  ax2-\-bx 

=p2x2,  which  gives  x=  - ,  which  value  put  in  the  other 

expression,  and  we   have   c(^^)  +\^ZZ^)==D' 

Multiplying  this  by  the  square  (p2 — a)2,  and  the  expres- 
sion becomes  cb2 — dbd-\-abp2=  some  square,  from  which 
the  value  of  p  can  be  found  and  afterwards  x. 


y 

And 


DIOPHANTINE   ANALYSIS.  IQg 

Example. 

A  certain  number  added  to  its  square,  the  sum  is  a  square, 
and  the  number  subtracted  from  its  square,  the  remainder  is 
a  square.     What  is  the  number? 
Let  x  =  the  number. 

Then  x2-{-x=Q.     And  x2 — #==  some  other  square. 

Assume  x=—  •    Then  —  (  \-\-y  )  =  Q. 

The  problem  will  be  solved  if  we  can  find  such  a  value 
of  y,  as  will  at  the  same  time  make  \-\-y  and  1 — y  squares. 

Therefore  put  l-^-y=p2,  and  1 — y=q2- 

From  the  first,  y=p2 — 1.     And  y=l — q2. 

Therefore,  q2=2 — p2.  As  q2  is  a  square,  we  have  only 
to  find  such  a  value  of  p2,  as  shall  render  2 — p2  a  square. 
But  this  cannot  be  done  unless  we  can  find  some  simple 
value  of  p  by  inspection,  and  we  do  observe  it  must  be  one. 
But  p  being  equal  to  one,  gives  y=0,  which  will  not  answer 
the  conditions.     Therefore,  let  p=l-\-t. 

Then  2— p2  =  \— 2t— t2={\—  ut)2  =  l—  2ut+u2t*. 

0r>fa^pT'  Take«=*  <_}  p=l+t=\. 
y=p2 — 1=| j-  x= — •  x  =  ||,  a  number  that  will  an- 
swer the  given  conditions. 

Case  3d.  To  resolve  a  triple  equality. 

Equations  in  the  form  of  ax  -\-  by  =  t2,  ax-\-dy=*  u*i 
ez-{-fy=s2,  can  be  resolved  thus  : 

■r>  •  c  j         dt2 — ^2 

By  expunging  ?/,  we  find  x=  — j — - — 

^2^,2 ct2 

Then  by  expunging  x.  y=-~- — — - . 

Substituting  these  values  of  a;  end  y  in  the  tnird  equation 


110  KEY  TO  ALGEBRA. 

j         i.  ni.         (af— bc)u2+(de— cf)t2       , 

and  we  shall  have  ^ '- — P~\ J—L—  =s2. 

ad — be 

Assume  «=±te.     Then  u-=t2z2.      Put  this  value  of 
u2  in  the  above,  and  divide  by  f,  and  we  shall  have 
(af — bc)z2-\-de — cf__s2 
ad — be  ~~  t*' 

The  right  hand  side  of  this  equation  is  a  square,  and 
therefore  all  that  is  now  requisite,  is  to  find  such  a  value  of 
z  as  shall  make  the  other  side  a  square,  which  when  possi- 
ble, can  be  done  by  case  7,  section  20. 

After  z  is  found  t  may  be  assumed  of  any  convenient 
value  whatever.  Now  u  is  known,  and  with  t  and  u  known 
quantities,  we  know  x  and  y. 

The  preceding  are  some  of  the  most  comprehensive  and 
general  methods  yet  known  ;  but  there  are  cases  in  practice 
where  no  general  rules  will  be  so  effectual,  as  the  operator's 
own  judgment  and  penetration. 

Much,  very  much  will  depend  on  skill  and  foresight  dis- 
played at  the  commencement  of  a  problem,  by  assuming 
convenient  expressions  to  satisfy  one  or  two  conditions  at 
once,  and  the  remaining  conditions  can  be  satisfied  by  som.6 
one  of  the  preceding  rules. 

Examples. 

1.  It  is  required  to  find  three  numbers  in  arithmetical 
progression,  such,  that  the  sum  of  every  two  of  them  may 
be  a  square. 

Let  x1  x-\-y  and  x-\-2y  represent  the  numbers. 

Then  by  the  general  formula, 

2x+y=t2,  2x-{-2y=u2,  2x-{-3i/=s2. 

-n  •      •  i         t* — V     u2 — % 

By  exterminating  x,  we  have  ■ — ^-'-= — jz — -• 

Continuing  thus  after  the  general  equations,  we*  find  a 


DIOPHANTINE   ANALYSIS.  HJ 

long  and  troublesome  process,  and  in  conclusion,  we  find  the 
numbers  to  be  482,  3362,  and  6242. 

The  above  is  according  to  the  common  as  well  as  the 
general  method. 

The  following  is  Mr.   Young1  s  Solution. 

Let  x — y,  a?,  and  x-\-y  represent  the  numbers.  Then 
2x—y,  2a?,  and  2x-\-y  must  be  squares. 

Assume  2x=m2-\-n2,  and  y=2mn. 

Then  2x — x=m2 — 2mn-\~n2,  and  2x-\-y=m2-\-2mn-\-n* 
are  evidently  squares.  It  therefore  only  remains  to  make 
2x  or  m2-\-n2  a  square,  and  this  can  be  done  as  explained  at 
the  close  of  the  last  section  by  assuming  m=r2 — 52,  and 
n=2rs.  Then  2x=.m2-\-n2=^{r2-\-s2Y^  an  expression  in 
which  r  and  s  can  be  assumed  in  numbers.  But  they  must 
be  so  assumed  that  x  shall  be  greater  than  y  to  make  x — y 
the  first  number,  positive  and  for  this  reason,  we  must  give 
the  literal  expressions  for  the  numbers  before  taking  definite 
values  for  r  and  s.  The  expressions  for  the  numbers  are 
S&— y=  £(r2+s2)2—  4rs  (r2— s2.)  x=  £(r2-f-52.)2 
x-\-y—  ±(r2-\-s2)2-\-4rs(r2=s2.) 

Take  r=9,  5=1,  and  482,  3362,  6242,  are  the  numbers. 

Another  Solution. 

x2  x2  x2 

Let  -^ y->~n     an(^    o — |-y  be  the  numbers. 

Then  x2 — y,  x2-\-y,  and  x2  must  be  squares. 

But  the  last  being  a  square,  we  have  only  to  make  x2 — y 
and  x2-\-y,  squares. 

Assume  y=2x — 1.  Then  x2 — y=x2 — 2#-|-l,  a  square, 
and  we  now  have  only  to  make  x2-\-2x — 1,  a  square. 

Therefore  make  *2-l-2# — \=(x-\-nf—x2-\-2nx-\-n2. 

X-2{l-n) 

It  is  manifest  that  n  must  be  less  than  one,  make  it  .& 

„,              1.64     41      _      x2     1681  72     1440 

Then  ... -=-    Or,  -  =  —  ^fo^W 


112  KEY  TO  ALGEBRA. 

m,       482    3362    6242 

Then400'     400 '     400"  are  the  numbers. 

Here  we  have  the  numbers  expressed  in  fractions,  but 
the  denominators  are  common,  and  is  a  square  number,  we 
may  therefore  multiply  all  three  by  400,  and  we  shall  have 
482,  3362,  and  6242  for  the  numbers,  as  in  the  other  solu- 
tions. 

If  we  take  n=  f  in  this  last  result,  we  shall  have  2162, 
7442,  and  9442  for  the  numbers. 

2.  Find  two  numbers  such,  that  if  to  each,  as  also  to  their 
sum,  a  given  square,  a2  be  added,  the  three  sums  shall  all 
be  squares. 

Let  x2 — a2  and  y2 — a2  represent  the  numbers :  then  the 
first  conditions  are  satisfied. 

It  now  remains  to  make  x2-\-y2 — 2a2-\-a2  a  square,  or, 
ff'+y2 — &2=n.  Assume  y2 — a2=2ax-\-a2.  This  assump- 
tion will  make  the  expression  a  square,  whatever  be  the 
values  of  either  x  or  a.  But  the  assumed  equation  gives 
y*=2&a+2&2,  and  as  y2  is  a  square,  we  must  find  such  values 
of  x  and  a,  as  shall  make  2ax-\-2a2:  a  square.  Put  x=na. 
Then  2na?-\-2a2=  □  ,or,  a2(2?i+2)=  □ .  Hence  it  is  sufficient 
that  we  put  2^+2  =  some  square.  Therefore,  assume 
2?i+2=16.  Hence  n=7  and  x=7a.  Now  take  a  equal 
to  any  number  whatever.  If  <z=l,  x=7,  ^=4,  and  48 
and  15  are  the  numbers,  add  1  to  each,  and  we  have  49  and 
16,  squares;  sum,  63+1=64,  a  square. 

3.  Find  three  square  numbers  whose  sum  shall  be  a 
square. 

Let  x2+y2+z2=  □ .  Assume  y2=2xz.  Then  x*+2xz-\-£ 
is  a  square.  But  2xz=\J.  Let  x=uz,  then  2wz2=D,  or 
2w=D=16,  tt=8,  x=8z,  z=l,  z=8,  y=4. 

Therefore  64+16+1  =81  =92. 

4.  Find  three  square  numbers  in  arithmetical  progression. 
Let  x2 — y,  x2.  and  x2-\-y  represent  the  numbers.    Assume 

x*=y2-\-\,then  the  first  and  last  will  be  squares,  and  it  only 


DIOPHANTINE  ANALYSIS.  ^3 

remains  to  make  the  middle  term,  or  y2+J>  a  square. 

©2 1 

Therefore,  put   y2-^-\z=(y — pyy  which  gives  y—~ 5— i' 

tip 

Take  ^=1,  then  #=£,  and  y2-\-  |=§f=#2.  Therefore, 
oVj  If?  if?  are  tne  numbers;  but  we  can  multiply  them  all 
by  the  same  square  number  64,  and  their  arithmetical  rela- 
tion will  not  be  changed,  and  they  will  still  be  squares  ; 
hence  1,  25,  and  49  may  be  the  numbers,  or  4, 100,  and  196, 

5.  Find  two  whole  numbers,  such  that  the  sum  and  dif- 
ference of  their  squares,  when  diminished  by  unity,  shall  be 
a  square. 

Let  #-f-l=one  number,  and  y=  the  other.  Then  by 
the  conditions  we  must  make  squares  of  x2-\-y2-*r2x,  and 
K2 — ^2_|_2#.  Assume  2x=a2,  and  y2=2ax,  then  the  ex- 
pressions become  x2-\-2ax-\-a2,  and  x2 — 2ax-\-a2,  obvious 
squares,  whatever  be  the  values  of  x  and  a.  But  the  equa- 
tions 2x=a2  and  y2=2ax  must  be  satisfied.  Take  a=4, 
then  a?=8,  y=S,  and  #-(-1=9.  Therefore  9  and  8  are  the 
numbers. 

6.  Find  three  whole  numbers,  such,  that  if  to  the  square 
of  each,  the  product  of  the  other  two  be  added,  the  three  sums 
shall  be  squares. 

Let  z,  xy,  xv:  be  the  numbers.  Then  by  the  conditions, 
x2-\-x2yv,  x2y2-\-x2v,  x2v2-\-z2y,  must  be  squares.  As 
each  term  contains  a  square  factor  #2,  it  will  be  sufficient  to 
make  l-{-yv=n1  y2-\-v=D1  and  v2-\-y=n> 

Assume  i/=4v-j-4,  and  this  will  make  the  first  and  last 
expressions  squares.  Substitute  this  value  of  y  in  the  second 
expression,  and  we  shall  have  16v2-j-33r-|-16,  which  must 
be  made  a  square.     Hence  put  16^2-)-33v-|-16=(4 — fv)2, 

which  reduced   gives  v=  2     JL    Take  ^=5,  then  v=7^ 

Now  take  #=9,  and  we  have  9,  73,  and  328  for  the  numbers. 

7.  Find  two  whole  numbers  whose  sum  shall  be  an  inte 

8 


H4:  KEY  TO  ALGEBRA. 

gral  cube,  and  the  sum  of  their  squares  increased  by  thrice 
their  sum  shall  be  an  integral  square. 

Let  x-\-y=n3,  that  is  some  cube.  Then  x2-\-y2-\-3n3=z 
□  .  Put  2xy=3n3,  then  x2  -\-2xy-\-y2  is  a  square,  whatever 
may  be  the  values  of  x  and  y.  But  x  and  y  must  conform 
to  the  equations  x-{-y=n3,  and  2xy=3n3.  Work  out  the 
value  of  x  from  these  equations,  on  the  supposition  that  n  is 
known,  and  we  shall  find  2x=?i3-\-tJ  \n6 — 6ft3.} 

Now  x  will  be  rational,  provided  we  can  find  such  a  value 
of  n  as  shall  render  n6 — 6n3  a  square,  but  if  we  add  9  to 
this,  we  perceive  it  must  be  a  square,  and  we  have  two 
squares,  which  differ  by  9.  Therefore  one  must  be  16,  the 
other  25,  as  these  are  the  only  two  integral  squares  which 
differ  by  9.  Hence  n6— 6?i3+9  =  25.  Or,  ?i3—- 3  ==  5. 
n3=8,  7i=2,  and  #==6,  y=2. 

8.  Find  three  numbers,  such  that  their  sums,  and  also  the 
sum  of  every  two  of  them,  may  all  be  squares. 

Let  x2 — 4a;  =  the  first,  4x=  second,  and  2#-|-l=  third. 
By  this  notation,  all  the  conditions  will  be  satisfied,  except 
the  sum  of  the  last  two.  That  is  Qx-\-l  must  be  a  square, 
but  fro  have  three  differe?it  whole  numbers,  no  square  will 
answer  under  121,  the  square  of  11.  Hence  put  6x-\-l  =  l21. 
Or,  a?=20.     And  the  numbers  will  be  320,  80,  and  41. 

9.  Find  two  numbers,  such  that  their  difference  may  be 
equal  to  the  difference  of  their  squares,  and  the  sum  of  their 
squares  shall  be  a  square  number. 

Let  x  and  y  be  the  numbers.  Then  x — y=x2 — y2. 
Divide  by  # — y,  and  l  —  x-\-y.  Hence  a?  =  l — y,  and 
z2-f-y2=l — 2j/-f-2y2.  Which  last  expression  1 — 2y-\-2y2 
must  be  made  a  square.     For  this  purpose,  put 

1— %+2#2=(l— nyf.     Hence  y=-J|^  • 

Take  n  any  value  to  render  y  less  than  one  in  order  to 


DIOPHANTINE   ANALYSIS.  ^5 

give  x  a  positive  value.      Therefore  take  w=3,  and  y=z  4 

Consequently  x=  f ,  answer. 

10.  Find  three  numbers  in  geometrical  progression,  such 
that  if  the  mean  be  added  to  each  of  the  extremes,  the  sums 
in  both  cases  shall  be  squares.  Ans.  5,  20,  and  80. 

11.  Find  three  numbers,  such,  that  their  product  increased 
by  unity  shall  be  a  square,  also  the  product  of  any  two  in- 
creased by  unity,  shall  be  a  square.  Ans.  1,  3,  and  8. 

Assume  1  for  the  first  number,  and  x  and  y  for  the  other 

12.  Find  two  numbers,  such  that  if  the  square  of  each  bo 
added  to  their  product,  the  sums  shall  be  both  squares. 

Ans.  9  and  16. 

13.  Find  three  integral  square  numbers  in  harmonica! 
proportion.  Ans.  25,  49,  and  1235. 

14.  Find  two  numbers  in  the  proportion  of  8  to  15,  and 
such  that  the  sum  of  their  squares  shall  be  a  square  number. 

Ans.  136  and  255.     Bonnycastle's  answer,  476  and  1080. 

15.  Find  two  numbers  such  that  if  each  of  them  be  added 
to  their  product,  the  sums  shall  be  both  square.    , 

Ans.  i  and  J. 
The  above  require  no  explanation  from  us. 

There  are  many  severe  and  tedious  problems  in  the  Diophan- 
tine  Analysis,  proposed  by  Bonnycastle,  Young,  and  others, 
which  require  more  time  and  practice  than  algebraists  in 
general  ought  to  give  for  the  advantage  derived,  as  time  and 
thought  may  be  better  employed  in  Analytical  Geometry, 
the  Calculus,  or  Astronomy. 


116  KEY  TO  ALGEBRA. 

THE  DIOPHANTINE  ANALYSIS. 

The  Diophantine  Analysis  is  sometimes  useful  in  solving 
numerical  Equations,  in  which  squares  and  cubes  are  in- 
volved, as  the  following  examples  will  show. 

1.  Given  j  ^__  _7  \  to  **n(*  one  va*ue  °^  x  an<*  V' 
By  subtraction,  and  the  transposition  of  y2  we  have 

x2+x-\-y=y2  (a) 

As  the  second  member  of  this  equation  is  a  square,  the 
first  must  be  a  square  in  fact,  if  not  in  form. 

But,  we  perceive,  that  if  we  put  y=x-\-\,  in  the  first 
member,  it  will  be  in  a  square  form. 

Put  this  value  of  y,  in  the  first  equation,  and  we  have 
x*-\-x=6;  which  gives  #=2;  hence  y=3;  values 
which  verify  both  equations. 

N.  B.  This  method  of  operation  is  not  general.  It 
only  serves  to  resolve  particular  cases.  We  might  have 
made  the  first  member  of  equation  (a),  a  square,  by  putting 
y—3x-\-4,  or  5#-f-9;  but  the  results  of  these  substitutions 
would  not  verify  the  primitive  equations. 

2.  Given  \  2x*—3xy+    2/2==4  ?  to  find  values  of  x  and  v. 

\    #2+3y2—  2xy=9  ) 

As  4  and  9  are  squares,  the  first  members  are  square  in 
fact,  though  not  in  form.  But  we  can  make  the  first  mem- 
bers square  in  form,  by  assuming 

2x2—3xy =0,     and  3y2 — 2#i/=0. 

Then  y2=4  and  #2=9,  or  i/=2  and  #— 3 ;  va- 
lues which  verify  all  the  equations. 


DIOPHANTINE   ANALYSIS.  117 

3.  Find  such  integral  values  of  x,  y,  and  z,  as  wiil  verify 
the  equations  ....  xi-\-yiJtxy=31i 

and  .  .  x2-)rzi-{-xz =49. 
If  we  add  xy  to  the  first  equation,  and  xz  to  the  second* 
the  first  members  will  be  square  ;  and,  of  course,  the  second 
members  will  be  square  in  fact,  though  not  in  form. 

We  have,  then,  to  make  37-f-#y>  and  4§-{-xzt  squares, 
*o  accomplish  this. 

Put         37-ftft/=49,       or      xy=12 (1) 

and     49~\-xz=64,      or       xz=\5 (2) 

12  15 

From  (1),  •  .  a?= — ;     from  (2) .  .  x=— . 

y  * 

Hence  «  •  12z=15y,  or,  z=— . 

Take  y=4,  then  z=5,  and  x=3 ;  values  which  will 
verify  the  given  equations. 

4.  Find  such  integral  values  of  y  and  z  as  shall  verify 
the  equation  ....  *y2-]-z2-\-yz=6l. 

Add  yz  to  both  members,  then  put    6l4-y*=n2. 

Now  if  we  assume n=8,  yz=3. 

But  yz—3  will  give  y-\-z=8,  and  these  two  equations 
will  not  give  integral  values  to  y  and  z.  Therefore,  take 
n=9,  then  w2=81,  t/z=20,  and  y-f-z=9.  Hence  z=4 
or  5,  and  y=5  or  4. 

5.  Find  such  values  of  x  and  y  as  will  verify  the  equa- 
tions   xy-\-xy2=l2  . (1) 

and.  ...  x  +xy*=\8 (2) 

12 

Equation  (1)  may  be  put  into  this  form  y*— — — y. .  .(3) 

*       # 
18 

Equation  (2)  into  this y8= 1...(4) 

x 

The  first  member  of  equation  (4)  is  a  cube ;  therefore 

Put 11-13=8.    Whence  z=2. 

x 


118  KEY  TO  ALGEBRA. 

r-  y-\-u-{-y2u2=^2l  [to  find  one  value  of  each 

|  x-\-v-\-2?v2  =41  |  of  the  symbols. 

A  regular  solution  would  result  in  a  very  high  and  tedi- 
ous equation ;  but  if  the  values  are  integral,  we  can  soon 
determine  them  as  follows  : 

Take  the  first  equation,  and  put  v-\-u=s,  and  transpose 
s.  Then  vhi2=l3 — s;  which  shows  that  13 — s  must  be 
some  square;  and  if  s  is  positive,  the  square  cannot  be 
greater  than  9.     That  is,  13 — #=9    or,  s=4. 

Then  v-f-w=4,  and  vu=3 ;  giving  v=l  or  3,  and 
t/=3  or  1.  By  taking  u=l9  in  the  second  equation,  we 
find  y=4.  With  the  values  already  found  we  obtain  x, 
from  either  the  third  or  fourth  equations,  v=3,  w=l, 
a?=2,  y=4. 

7.  Given   \  ^~~~       (  to  find  values  of  x  and  y. 

{  2?/2-f-3.ri/=  8  J  * 

Put  xy=p  ;    transpose,  &c,  we  have 

4#2=12-{-2;)    and    4i/2=16— 6p. 

Now  we  must  find  such  a  value  of  p  as  shall  render  the 
second  members  of  these  last  equations  square  at  the  same 
time;  which  is  p=2;    this  gives  4x*=  16,   x=2. 

ft    p.         C  6x2-\-2y2  =5xy -f- 12  )  to  find  one  value  of 

(  3x2-\-2xy=3yz  —  3  )  x  and  y. 
This  problem  is  under  (Art.  110)  of  both  editions. 
Ac(^  the  equations  together,  and  reduce,  and  we  have 

9x2=y2-\-3xy-{-9. 
The  first  member  of  this  equation  is  a  square ;  therefore  the 
second  member  is  a  square,  but  to  make  it  a  square  inform, 
as  well  as  in  fact,  we  perceive  it  is  only  necessary  to  make 


MISCELLANEOUS  EXAMPLES.  H9 

07=2.  Or  call  y2-\-3xy-\-9  a  binomial  square,  and  decide  the 
value  of  x  agreeable  to  section  8,  This  gives  a?=2,  the 
answer. 

q    p-         5  ^2 —  *'/=  28?  to  find  the  rational  values 
'  °1V       ^4y24-3ary=160  5  of  a?  and  y.  (Y.   139.) 

Ans.  x  =  ±4,  y=  ±5. 
There  are,  and  may  be  equations  of  the  preceding  forms 
which  have  no  rational  values,  such  are  not   susceptible  of 
of  this  mode  of  treatment. 


OF   THE 


\ 


Section  XXIIli  UNI  VERS  IT 


Miscellaneous  Exampl 


%IP0R^' 


1.  When  wheat  was  8  shillings  a  bushel  and  rye  5,  a 
man  wished  to  fill  his  sack  far  the  money  he  had  in  his 
purse.  Now  if  he  bought  15  bushels  of  wheat  and  laid  out 
the  rest  of  his  money  in  rye,  he  would  want  3  bushels  to 
fill  the  sack:  but  if  he  bought  15  bushels  of  rye,  and  then 
filled  his  sack  with  wheat,  he  would  have  15  shillings  left. 
How  much  of  each  must  he  purchase  to  fill  his  sack,  and 
lay  out  all  his  money  ? 

(Colburn,  page  50.)     Ans.  10  bushels  of  each. 

Solution  by  Mr.  T.  J.  Matthews. 

Let  x=  the  wheat,  and  y=  the  rye.  Then  it  is  evident 
that  when  he  buys  15  bushels  of  wheat  he  has  too  much,  as 
he  has  not  money  enough  left  to  fill  his  sack  with  rye.  Now 
15 — x  is  the  excess  of  the  wheat  purchased  above  what  he 
ought  to  have  had,  and  this  excess  of  quantity,  multiplied  by 
the  excess  of  a  bushel  of  wheat  above  one  of  rye,  wiJJ  give 
the  deficiency  of  his  money,  or  equal  to  3  bushels  of  rye  at 
5  shillings.     Consequently,  3(15 — a?) =15,  or  #=10. 

By  similar  reasoning,  it  will  appear  that  when  15  bushels 
of  rye  are  purchased,  he  buys  too  much,  and  the  excess  is 
15 — y,  which  multiplied  as  before,  will  equal  the  excess  of 


120  KEY  TO   ALGEBRA. 

his  money,  viz:  15  shillings.      Therefore  3(15 — y)=15, 
and  y= 10. 

2.  A  person  bought  two  cubical  stacks  of  hay,  for  ^41 ; 
each  of  which  cost  as  many  shillings  per  solid  yard  as  there 
were  yards  in  a  side  of  the  other,  and  the  greater  stood  on 
more  ground  than  the  less  by  9  square  yards.  What  was 
the  price  of  each  1  (Colburn.) 

Solution  by  T.  J.  Matthews. 
Assume  5z,  and  4z  the  sides  of  the  Cubes. 
Then  25a;3— 16^=9,  by  the  first  condition. 
Therefore,  x=h         1254=500.        64-5=320. 
3  Or,  £26.        £IQ.  Answer. 

3.  Given  x-\-y-{~z=25.  ary=6  #2=60,  to  find  #,  y 
and  z. 

Solution.  From  the  two  latter,  z—lOy.  Then  the  first 
becomes  x-\-lly—25.  Or  z2-\-Uxyz=2ox,  but  from  the  2d, 
ll2?y=66.     Hence  ^—-25^=— 66.  (A) 

Assume  2«= — 25,  then  6&-f-9= — 66,  and  equation  (A) 
becomes   (z2-{-2ax-{-di=a2-{-6a-{-9.     Or,  a?=3,) 

4.  Given  a^=125#-|-300y.  And  y2— x2 =90000,  to  find 
*  and  y.  (Young  p.  146.)     Ans.  #=400.     y=500. 

Put  y  =px,  then  px2 = 1 25#+300^#.         ( 1 ) 

And  p2y2—x2=(300)2  (2) 

125 
From  equation  (1)  p=  ~~^  (3) 

(300) 2 
From  equation  (2)  p2 — l  =  v — — 

THe  right  hand  side  of  this  last  equation  being  a  square 
the  other  side  is  also  a  square,  and  one  accustomed  to  the 
analysis  will  perceive  that  p  must  equal  f,  to  make  the  ex- 
pression p*  — 1  =  D-  Others  can  go  through  the  form  and 
they  will  find  that  p=l,  which  value  put  in  equation  (3) 
gives  a?a=400. 


MISCELLANEOUS  EXAMPLES.  121 

It  is  not  imperative  that  we  should  resort  to  the  Diaphan- 
tine  to  solve  this  problem  but  it  is  very  Convenient. 

5.  Find  two  numbers,  such  that  the  fifth  power  of  one 
may  be  to  the  cube  of  the  other,  as  972  to  125. 

Ans.  6  and  10. 

Let  a?  and  nx  be  the  numbers. 

Then  a;5  :  n3x3  :  :  972  :  125.  Or,  x2  :  n3  :  :  972  :  125. 

Multiply  the  first  and  third  terms  by  x,   x3  :  n3  :  :  972a: 

:   125. 

1 25x3 
Therefore,  972*=— f-. 
n3 

The  right  hand  side  of  this  equation  is  a  cube,  therefore, 
972#=  a  cube:  Or,  27*36x=a  cube.  Hence,  36a?  must  be 
a  cube,  which  it  evidently  is,  when  #=6,  as  36  is  the  square 
of  6. 

6.  Given  x+i,+xy(x+y)+r* y2 =85    )      fi   ,  , 
And     xy+{x+?jy+xy(x-\-y)=97  \  V' 

Young.  145.     Ans.  x=Q  y=l. 
Put  (x-\-y)=s  xy=p,  then  the  equations  become 

s-\-sp+p2  =85  } 
And  p-\-sp-\-s2  =97  $ 

By  addition  (s-\-p)+s- -\-2sp-\-p2  =\82.  (1) 

Assume  Q,=s-{-/>,  then  equation  (1)  becomes 
Q,2_|_a==18o       Hence  Q=  13.     Or,  s+p=l3.     The  re- 
maining steps  are  obvious. 

12 

7.  Given  lJx-\-l2=    — -=,  to  find  x. 

144 
Square  x-{- 12=^=  -£-.     Put  .r-f-5=y. 

144 

Then  y+7= .     Or,  ^+7^=141 

Put  2&=7.  Then  18a-f  81  =  144. 
Hence  y>+2ay+a*=a2+l8a~)-8h 
y-{-a=a-(-9.     Or,  — a— 9. 


122  KEY  TO   ALGEBRA. 

&  Given   J  (a?+y)(«y+l)=18^-  l  to  find  z  and 

o.  driven,  j  (a?2_jly2>)^2?/2_^1)=208a,2y2   J  y 

Ans.  o:=2±V3-     0r  ?±V3- 
y=7=fc4«/3.     Or2d=V3. 

Solution.  Take  x-\-y=s.     xy-\-l  =  t and  xy=p. 

Thenst=lSp.     (1)  And  (s9—2p)(t2—2p)=20Sp2.     (2) 

Multiply  as  indicated  and  take   the   value  of  s't2  from 

equation  (1)  and  we  have  324/— 2p  (s2+22)+4/=208/. 

„         ,      .  **4-*a  .  st 

By  reduction,  p=  —  '  - — .     From  equation  (1)  p=z—^. 

Therefore,     ~*~    =-.     Assume  s—nt. 

00         lo 

Then  this  last  equation  becomes,  by  a  little  reduction, 
n2+l      »      „  0    '"      1 

HO    =3-     Hencew=3-     0r'3- 

This  establishes  a  relation  between  x  -f-  y  and  #y-|-L 

Another  Solution  by  Charles  E.  Matthews. 

Multiply  the  original  equations  as  indicated,  and 

x  2y-\-xy  -  -\-x-\-y=  1 8xy. 

x*y2-{-x2yA^-x2-\-y2=2Q8x2y2. 

Divide  the  first  by  xy,  the  2d  by  x2y2y  then 

x+y+l  +1 =ia  And  x2+y*+^+}*=208- 

Assume   x-\ — =??i.    And  y-A — =?j. 
^x  s^y 

Then  w+w=18.     And  m2-\-n2  =212. 

From  which  »i  and  ?i   are  easily  found,  afterwards  x 

and  y. 

9.  A  square  public  green  is  surrounded  by  a  street  of  uni- 
form breadth.  The  side  of  the  square  is  3  rods  less  than  9 
times  the  breadth  of  the  street:  and  the  number  of  square 
rods  in  the  street  exceeds  the  number  of  rods  in  the  perime- 
ter of  the  square  by  228.     What  is  the  area  of  the  square  ? 

(Day  307.)     Ans.  576  rods. 

10.  A  man  wishes  to  purchase  a  certain  number  of  acres 


MISCELLANEOUS  EXAMPLES.  1x3 

of  land  for  the  money  he  has  at  his  command.  Cleared  land 
is  worth  10  dollars  per  acre;  uncleared  land  is  worth  $8.-— 
He  finds  that  if  he  buys  120  acres  of  cleared  land,  and  lays 
out  the  rest  of  his  money  for  that  which  is  not  cleared,  he 
will  not  get  the  quantity  of  land  he  wants  by  25  acres,  but, 
if  he  buys  220  acres  of  uncleared  land,  and  then  buys  a  suf- 
ficient number  of  acres  of  cleared  land  to  make  up  the  num- 
ber of  acres  he  wants,  he  will  have  4  dollars  left.  How 
many  acres  of  each  must  he  buy  to  have  the  quantity  he 
wishes,  and  lay  out  all  his  money?  (Harney  page  203. 
Ans.  20  acres  cleared,  218  uncleared. 
N.  B.  Call  to  mind  problem  first  of  this  section. 


In  working  geometrical  problems  algebraically  much  la- 
bor may  be  saved  by  paying  attention  to  the  relation  oi  the 
given  numbers. 

We  give  the  following  as  illustrative  of  these  remarks. 

1.  If  the  perimiter  of  a  right  angled  triangle  be  720,  and 
the  perpendicular  falling  from  the  right  angle  on  the  hy- 
pothenuse  be  144 ;  what  are  the  lengths  of  the  sides  ? 

(Day  Alg.  p.  305.)     Ans.  300,  240  and  180. 

If  we  use  the  identical  numbers  given  144  and  720,  as 
nine-tenths  of  our  teachers  do,  they  will  give  large  and  tedi- 
ous equations,  but  if  we  compare  144  and  720,  we  shall 
perceive  that  one  is  exactly  5  times  the  other,  and  consider- 
ing the  nature  of  similar  triangles,  we  can  work  on  one  of 
only  144th  of  the  linear  dimensions  of  the  first,  or  a  triangle 
whose  perimiter  is  5,  and  perpendicular  from  the  right 
angle  1. 

Solution.  Let  x  and  y  be  the  two  sides,  then  5 — x — y  will 
be  the  hypothenuss. 

And  x*-\-y*=(o — x — y)?,  and  xy=5 — x — y. 

Each  member  of  this  last  equation  expresses  the  double 
area  of  the  triangle.     Put  x-\-y=s.     xy=p. 


124  KEY  TO  ALGEBRA. 

Then  $2_2/?=(5— s)2=25— 10s+s2,  and ^=5— *. 
Or,  105  —2^=25 
And  25  +2j5=10 


By  addit'n  12s  =35.      _  35 

0r'5=12' 


But  s=x-\-y,  the  sum  of  the  two  sides  which,  taken 

from    5,   or    — >,  gives— ,  for  the  hypothenuse  of  the  small 

25 

triangle,  hence      by  144=300,  the  hypothenuse  of  the  large 

triangle. 

2.  The  sum  of  the  two  sides  of  a  plane  triangle  is  1155, 
the   perpendicular  drawn  from  the  angle  included  by  these 
sides  to  the  base,  is  300 ;  the  difference  of  the  segments  of 
the  base  is  495,  what  are  the  length  of  the  three  sides  ? 
(Day  305.)     Ans.  945,  375,  780. 

Write  the  given  members  in  order,  thus  300,  495,  1155. 
Divide  them  by  15,  and  their  relation  is  20,  33,  77. 

The  two  latter  numbers  have  a  common  factor  11,  which 
call  a.     Put  6=20. 

Then  the  three  given  lines  will  be  b,  3a,  and  la.  Let  x= 
the  less  side,  la — z=the  greater  side,  #=the  shorter  seg- 
ment, and  g-[-3a—the  longer  segment  of  the  base. 

Then  p*-\-b*=xx*.  (1 

And  y2-\-Qa?j-\-9a2-\-b2=49a2^Uax+x2.        (2) 

Subtract  (1)  from  (2),  drop  9a2  from  both  sides,  and  di 
vide  by  2a,  and  3y=ab — Ix.     (3) 

We  write  2b  in  place  of  40,  after  dropping  9a2. 

From  the  square  of  (3)  subtract  9  times,  equation  (l,)and 
we  have — 9b2=a2b3 — 14  abx-\-2bx*. 

Divide  by  b,  afterwards  by  2,  recollecting  that  6=20,  and 

we  have  — 90=10&2 — lax-\-x2. 

9a2 
Add  —j-,  to  both  sides  to  complete  the  square. 


MISCELLANEOUS  EXAMPLES.  125 

Then?|— 90=  ®  (a»-40)==®-81  =  ^->tax+x>. 

Extract  square  root- "9=     — x. 

Or,  a?=25.     Then  25. 15=375. 

3.  Divide  the  number  74  into  two  such  parts  that  the 
difference  of  the  square  roots  of  the  parts  may  be  2. 

Ans.  25  and  49. 
Let  x — 1 ,  and  x-\-\  be  the  square  roots,  of  the  two  parts. 
This  problem  can  also  be  solved  by  the  Diophantine  analysis. 

4.  Given  £*+#2=45  and— -) — =£,  to  be  solved  by  the 

x      y 

Diophantine  analysis.  Ans.  x—6.     y=3. 

5  Given  a?2-f-#2=45  and  (x-\-y)x=54  to  find  x  and  y  by 
the  Diophantine  analysis.  Ans.  x=Q.     y=3. 

The  two  preceding  should  also  be  worked  by  common 
algebra. 

6.  A  and  B  traveled  on  the  same  road,  and  at  the  same 
rate,  from  Huntingdon  to  London.  At  the  50th  mile  stone 
from  London,  A  overtook  a  drove  of  Geese,  which  were 
proceeding  at  the  rate  of  3  miles  in  2  hours,  he  afterwards  met 
a  stage  wagon,  which  was  moving  at  the  rate  of  9  miles  in 
4  hours.  B  overtook  the  same  drove  of  Geese  at  the  45th 
mile  stone,  and  met  the  same  stage  wagon  exactly  forty  mi- 
nutes before  he  came  to  the  31st  mrle  stone.  Where  was  B 
when  A  reached  London  1 

Solution.  Let  #=  miles  traveled  by  each  per  hour,  and 
y=  distance  B  was  behind  A,  then  50 — 2x=  the  distance 
from  London  where  A  met  the  wagon. 

Also,~-Jg__  —  —   time  elapsed  between  the  meeting 

9y 
of  the  wagon  with  A  and  B,  therefore  -.—- ?-x=  distance  tra- 

4lr-f-9 

veled  by  the  wagon  during  this  time,  consequently 

9?/ 
50 — %x-\-d  _.    =distance  of  the  wagon  from  London,when 


126  KEY  TO   ALGEBRA. 

2a? 
met  by  B.     But  this  distance  is  also  =31-}-  ~  therefore 

0 

50_2x+4-^=31+|:.      Again  y+5  :  5  : :  »  :  |, 

whence  3y-|-15=10a?  and  y= ,   substituting    and 

reducing,  we  get  the  quadratic 

.      123a:     378      xktL 

x2 =  -s— .     Whence  a?=9,  consequently  y=25. 

7.  Given  a?2-f-a?y=77,  and  xy — y2  =  12,  to  find  a?  and  y, 
by  the  Diophantine  analysis.  Ans.  x—7.     y=4:. 

8.  Three  equal  circles  touch  each  other  externally,  and 
enclose  between  the  points  of  contact  a  acres  of  ground, 
what  are  the  radii  of  the  circles  ? 

AKo,i6ik)i 

9.  A  person  has  £21  65.  in  guineas  and  crown  pieces,  out  of 
which  he  pays  a  debt  of  £14  17s.,  and  finds  that  he  has  ex- 
actly as  many  guineas  left  as  he  has  paid  crowns  away ; 
and  as  many  crowns  as  he  has  paid  away  guineas ;  how 
many  of  each  had  he  at  first  ? 

Ans.  9  crowns  paid  away ;  12  guineas  paid  away. 
Suppose  a?=  the  guineas  paid  away. 
And        y=-  the  crowns  paid  away. 
Then  21a?-|-5#=297=  amount  paid  out  {  . 

And  5a?  +21^=249=  amount  on  hand  \  Per  <luestlon- 
Add  these  equations  and  divide  by  26,  &c. 

10.  The  sum  of  three  numbers  in  harmonical  proportion  is 
191,  and  the  product  of  the  first  and  third  is  4032.  What 
are  the  numbers  ?  Ans.  72,  63,  56. 

11.  Is  it  possible  to  pay  £50  by  means  of  guineas  and 
three  shilling  pieces  only.  Ans.  Impossible. 

12.  A  merchant  drew  every  year  upon  the  stock  he  had  in 
trade,  the  sum  of  a  dollars  for  the  expense  of  his  family. 

His  profits  each  year,  were  the  nth  part  of  what  remain- 


10.  Given  \  V(5V*+5V</)+V*+Vy=l<>  )    to  find 


MISCELLANEOUS  EXAMPLES.  127 

ed  after  this  reduction,  but  at  the  end  of  the  3d  year  he  finds 

his  stock  exhausted;  how  much  had  he  at  the  beginning; 

a(3n2+3n+l) 

Ans. . 

(n+1)2 

And  a?*+y*=s275  )  x  and  V* 

Put  Jipjx+bjyj^n.  Then  the  first  equation  becomes 

n2 

.-- \-n=\0.     Which  equation  gives  n=5. 

Whence  Jx-\-,Jy=5. 

From  this  last  and  the  2d  equation,  we  find  x=9,  and 
t/=4. 

The  following  are  from  Bland's  Problems,  and  involve 

equations  only  of  the  second  degree.     They  are  too  severe 

for  learners,  but  we  are  tempted  to  leave  one  or  two  of  them, 

without  solution,  for  the  benefit  of  those  who  deem  keys 

unnecessary.     More  of  like  character  might  be  given. 

18  7 

x = .  Ans.     #=16   or   1. 

*     Jx— 2 

2x*(x*-\-cciy=2xi(x-\-2a)-\-a2(x— a). 

Ans.  x=\a    or  — a. 

*-f(«V—  n+2*)*  =7+2^- y* ; 

(3i/  —  #+7  )*  =  ^.         Ans.  *=4,  y=2, 
x—y 

Double  the  first  equation,  transpose  — 2y2,  and  subtract  1 1 
from  both  members,  then  we  have 

2y2+2x—  ll+2(3i/2—  ll+2#)^=3+4y. 
Add  yl  to  both  members,  and  conceive  the  terms  in  the  vin- 
culum to  be  P  ;  then 

P2+2P=3+4y+y«; 
Or,         P+l=2+y. 


128  KEY  TO   ALGEBRA. 

By  restoring  the  value  of  P,  and  reducing,  we  have 
a?=6-f-2/ — y2.         Put  this  in  the  2d  eq.,  &c. 
i 
4.  Given  x2y — 4=4x*y — |?/3, 

and  x2 — 3=#y  (  xa — y*  J,  to  find  x  and  y. 

Ans.    fl?=l,  y=4. 


Put  #  =P,     ya=Q,    and  we  have 

P4Q2__4  =  4PQ2_|Q61 (!) 

P3_3     =PQ(P— Q), (2) 

Now  put  P=nQ,  and  eq.  (1)  becomes 
(4rt4-r-l)Q6—  16rcQ3=16. 
Conceive  n  to  be  a  known  quantity,  then  the  last  equa- 
tion is  quadratic,  and  a  solution  gives 

4(2tt2-r-2tt-r-l)_  4  _  4 

4n4-p-l  2m2— 2/i-f-l      2n*— 2n-f2—  1* 

But  from   (2),    Q3=  3     32  ,     =-— L-— 

Put  the  two  values  of  Q3  equal,  and  put  n2 — n-f-l=R,  (3) 

Th6n  2^-I=i     Whence  2„=6|=? (4) 

But  from  (3),  resolved  as  a  quadratic, 

2n=l±V4R^3 (5) 

From  (4)  and  (5),  2R±2R^/4R~ 3~=6R— 3 ; 
Or ±2R74R— 3=4R— 3. 


Put V4R— 3=S, 

Then  S2±2RS=0,         Or,        S(S=b2R)=0. 

This  last  equation  may  be  verified  by  taking  either  factor 
equal  to  zero ;  and  as  the  first  factor  only  gives  a  rational 
quantity,  we  take  that  which  gives  R=f . 

By  retracing,  we  easily  find  x  and  y. 


TECS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 

AN  INITIAL  PINE  OF  25  CENTS 

WILL  BE  ASSESSED   FOR   FAILURE  TO   RETURN 
THIS   BOOK  ON   THE   DATE  DUE.   THE   PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY    AND    TO    $I.OO    ON    THE    SEVENTH     DAY 
OVERDUE. 

EC  26  194f 

JUL  22  1942 

AUG   5  W& 

FEB    S  1947 

LD  21-100m-7,'40 (6936s) 

\  5-3 

UNIVERSITY  OF  CALIFORNIA  LIBRARY 


